Solutions to Exercises from Kenneth Brown’s …cgerig/solns.pdfSolutions to Exercises from Kenneth Brown’s Cohomology of Groups Christopher A. Gerig, Cornell University (College

Solutions to Exercises from

Kenneth Brown’s Cohomology of Groups

Christopher A. Gerig, Cornell University (College of Engineering)

August 2008 - May 2009

I appreciate emails concerning any errors/corrections: [email protected]. Anyerrors would be due to solely myself, or at least the undergraduate-version of myself

when I last looked over this. Remark made on 1/28/14.

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major: Applied & Engineering Physicsyear: Sophomore Undergraduate

Hey Ken, thanks for taking me under your wing to learn math.

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Contents

0 Errata to Cohomology of Groups 4

1 Chapter I: Some Homological Algebra 5

2 Chapter II: The Homology of a Group 12

3 Chapter III: Homology and Cohomologywith Coefficients 21

4 Chapter IV: Low-Dimensional Cohomologyand Group Extensions 31

5 Chapter V: Products 41

6 Chapter VI: Cohomology Theoryof Finite Groups 47

7 Chapter VII: Equivariant Homology andSpectral Sequences 55

8 Chapter VIII: Finiteness Conditions 58

9 Chapter IX: Euler Characteristics 59

10 Additional Exercises 61

11 References 76

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0 Errata to Cohomology of Groups

pg62, line 11 missing a paranthesis ) at the end.pg67, line 15 from bottom missing word, should say “as an abelian group”.pg71, last line of Exercise 4 hint should be on a new line (for whole exercise).pg85, line 9 from bottom incorrect function, should be

∑g∈C/H g

−1 ⊗ gm.pg114, first line of Exercise 4 misspelled endomorphism with an extra r.pg115, line 5 from bottom missing word, should say “4.4 is a chain map”.pg141, line 4 from bottom missing a hat on the last H∗.pg149, line 13 the ideal I should be italicized.pg158, line 20 there should be a space between SL2(Fp) and (p odd).

pg160, line 5 from bottom missing the coefficient in cohomology, H∗(H,M).pg165, line 26 the comma in the homology should be lowered, Hq(C∗,p).

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1 Chapter I: Some Homological Algebra

2.1(a): The set A = g − 1 | g ∈ G, g 6= 1 is linearly independent because∑αg(g− 1) = 0⇒

∑αgg =∑

αg =∑αg1 +

∑0g, and since ZG is a free Z-module,

∑αg has a unique expression, yielding

αg = 0 ∀g ∈ A. To show that I = span(A), let∑αgg ∈ I and hence

∑αg = 0. Thus we can write∑

αgg =∑αgg − 0 =

∑αgg −

∑αg =

∑αg(g − 1). Since A is a linearly independent set which

generates I, it is a basis for I.

2.1(b): Consider the left ideal A = (s− 1 | s ∈ S) over ZG.A ⊆ I since ε(A) =

∑s ε((∑g rgg)(s− 1)

)=∑s ε(∑g rgg)ε(s− 1) = y · 0 = 0.

If x ∈ I then x =∑αigi −

∑βjgj such that

∑αi =

∑βj .

x = (∑αigi −

∑αi)− (

∑βjgj −

∑βj) + (

∑αi −

∑βj)

Thus x =∑αi(gi − 1)−

∑βj(gj − 1) and it suffices to show g− 1 ∈ A for any g ∈ G so that x ∈ A and

I ⊆ A. Since G = 〈S〉, we have a representation x = s±11 · · · s±1

n . By using ab − 1 = a(b − 1) + (a − 1)and c−1 − 1 = −c−1(c− 1), the result follows immediately.

2.1(c): Suppose S ⊂ G | I = (s− 1 | s ∈ S). Then every element of I is a sum of elements of

the form g − g′ | g = g′s±1. For g ∈ G, g − 1 ∈ I, so we have a finite sum g − 1 =∑n−1i=1 (gi − g′i).

Since this is a sum of elements in G where G is written multiplicatively, ∃ i0 | gi0 = g, say i0 = 1.

Thus g − 1 = g − g′1 +∑n−1i=2 (gi − g′i) ⇒ g′1 − 1 =

∑n−1i=2 (gi − g′i). Another iteration yields g2 = g′1 and

g = g1 = g′1s±11 = g2s

±11 = g′2s

±12 s±1

1 . At the last iteration, g′n−2 − 1 = gn−1 − g′n−1 ⇒ g′n−1 = 1, g′n−2 =

gn−1 = g′n−1s±1n−1 = 1s±1

n−1. Through this method we obtain a sequence g1, g2, ..., gn−1, gn | gi = gi+1s±1i

and g1 = g, gn = 1. ∴ g has a representation in terms of elements of S and so G = 〈S〉 since g wasarbitrary.

2.1(d): If G is finitely generated, then by part(b) above, I is finitely generated. For the converse,suppose I = (a1, ..., an) is a left ideal over ZG. Noting from part(a) that I = 〈g − 1〉g∈G as a Z-module,each ai can be represented as a finite sum ai =

∑j zj(gj−1). Since each ai is generated by finitely many

elements, and there are finitely many ai, I is finitely generated as a left ideal by elements s − 1 wheres ∈ G. Therefore, we apply part(c) to have G = 〈s1, . . . , sk〉 and thus G is a finitely generated group.

2.2: Let G = 〈t〉 with |G| = n and let t be the image of T in R = Z[T ]/(Tn − 1) ∼= ZG. The el-ement T − 1 is prime in Z[T ] because Z[T ]/(T − 1) ∼= Z which is an integral domain (An ideal P ofa commutative ring R is prime iff the quotient ring R/P is an integral domain; Proposition 7.4.13[2]).By Proposition 8.3.10[2] (In an integral domain a prime element is always irreducible), T − 1 is irre-ducible in Z[T ]. We also could have obtained this result by applying Eisenstein’s Criterion with thesubstitution T = x+ 3 and using the prime 2. Since Z[T ] is a Unique Factorization Domain, the specificfactorization Tn − 1 = (T − 1)(Tn−1 + Tn−2 + · · ·+ T + 1) with the irreducible T − 1 factor is unique,

considering the latter factor∑n−1i=0 T

i as the expansion of its irreducible factors [note: if n is prime then∑n−1i=0 T

i = Φn(T ), a cyclotomic polynomial which is irreducible in Z[T ] by Theorem 13.6.41[2]]. Thus

every f ∈ R is annihilated by t− 1 iff it is divisible by N =∑n−1i=0 t

i (and vice versa), and so the desiredfree resolution of M = Z = ZG/(t− 1) is:

· · · → Rt−1−→ R

N−→ Rt−1−→ R→M → 0 .

3.1: The right cosets Hgi are H-orbits of G with the H-action as group multiplication. Since G =⊔Hgi

where gi ranges over E, ZG =⊕Z[Hgi] ∼=

⊕Z[H/Hgi ]. G is a free H-set because hg = g ⇒ hgg−1 =

gg−1 ⇒ h = 1, i.e. the isotropy groups Hg are trivial. Therefore, ZG is a free ZH-module with basis E.

3.2: Let 〈S〉 = H ⊆ G and consider Z[G/H]. Now x ∈ Z[G/H] has the expression x =∑zi(giH),

and there exists an element fixed by H, namely, x0 = g0H = H where g0 ∈ H. H is annihilated byI = Kerε = (s− 1) since (s−1)H = sH−H = H−H = 0 ∀s ∈ S, and so Ix0 = 0. We have g−1 ∈ Isince ε(g − 1) = ε(g)− ε(1) = 1− 1 = 0. Hence (g − 1)x0 = 0⇒ gx0 = x0 ∀g ∈ G⇒ Gx0 = x0. Finally,GH = H ⇒ G ⊆ H. ∴ G = H = 〈S〉.

4.1: Orienting each n-cell en gives a basis for Cn(X). If X is an arbitrary G-complex, then with

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∑ηie

ni ∈ Cn(X), g ∈ G can reverse the orientation of en by inversion (fixing the cell). Thus G need not

permute the basis, and hence Cn(X) is not necessarily a permutation module.

4.2: Since X is a free G-complex, it is necessarily a Hausdorff space with no fixed points under theG-action. First, assume G is finite and take the set of elements in Gx0, which are distinct pointsgix0 with 1x0 = x0 for an arbitrary point x0 ∈ X. Applying the Hausdorff condition, we have opensets Ugi containing gix0 where each such set is disjoint from U1 containing x0. Form the intersectionW = (

⋂i g−1i Ugi) ∩ U1 which contains x0. Since gkW ⊆ Ugk, we have gkW ∩W = ∅ for all nonidentity

gk ∈ G, and so W is the desired open neighborhood of x0 [This result does not follow for arbitrary Gsince an infinite intersection of open sets need not be open].Assume the result has been proved for G infinite.Let ϕ : X → X/G be the quotient map, which sends the disjoint collection of giW ’s to ϕ(W ). Sinceϕ−1ϕ(W ) =

∐i giW , giW → ϕ(W ) is a bijective map (restriction of ϕ) and thus it is a homeomorphism

(ϕ continuous and open). This covering space is regular because G acts transitively on ϕ−1(Gx) by def-inition. Elements of G are obviously deck transformations since Ggx = Gx, hence G ⊆ Aut(X). GivenΓ ∈ Aut(X) with Γ(a) = b, those two points are mapped to the same orbit in X/G (since ϕ Γ = ϕ),and so ∃ g ∈ G sending a to b. By the Lifting Lemma (uniqueness) we have Γ = g, hence Aut(X) ⊆ G⇒G is the group of covering transformations.If X is contractible then G ∼= π1(X/G)/π1(X) ∼= π1(X/G)/0 = π1(X/G) and ϕ is the universal cover ofX/G, so X/G is a K(G, 1) with universal cover X.It suffices to show that the G-action is a “properly discontinuous” action on X when G is infinite:Every CW-complex with given characteristic maps fj,n : (Bnj , S

n−1j ) → (σnj , ∂σ

nj ) admits a canonical

open cover Uσ indexed by the cells, where Ua and Ub are disjoint open sets for distinct cells of equaldimension (for instance, if X is a simplicial complex we can take Uσ = St(σ) which is the open starof the barycenter of σ in the barycentric subdivision of X). More precisely, for one cell σnk ⊂ Xn ineach G-orbit of cells define its “barycenter” as σnk ≡ fk,n(bk), where bk = 0 ∈ Bn is the origin of then-disk; for the rest of the cells σni = giσ

nk in each G-orbit define their “barycenters” as σni ≡ fi,n(bi),

where bi ∈ Bni is chosen so that σni = giσnk . Considering the 0-skeleton X0, its cells σ0

j are by defini-

tion open and so the canonical open cover of X0 is the collection Uσ0j

= σ0j . Proceeding inductively

(with Uσ open in Xn−1), consider the n-skeleton Xn = Xn−1⋃j σ

nj and note that the preimage under

fj,n of the open cover of Xn−1 is an open cover of the unit circle Sn−1j . Take an open set f−1

j,n (Uσ)

in Sn−1j and form the “open sector” Wj,σ ⊂ Bnj which is the union of all line segments emanating

from bj and ending in f−1j,n (Uσ), minus bj and f−1

j,n (Uσ); each Uσ determines such a Wj,σ. As fj,n is ahomeomorphism of Int(Bnj ) with σnj , we have such open sectors fj,n(Wj,σ) in the n-cell. Noting theweak topology on X, the set U ′σ = Uσ

⋃j fj,n(Wj,σ) is open in Xn iff its complement in Xn is closed

iff (Xn − U ′σ) ∩ σik ≡ σik − (U ′σ ∩ σik) is closed in σik for all cells in Xn. For i < n, U ′σ ∩ σi = Uσ ∩ σibecause fj,n(Wj,σ) ⊂ σnj which is disjoint from the closure of all other cells, and the complement of this

intersection in σi ⊂ Xn−1 is closed because Uσ is open by inductive hypothesis. Therefore, it suffices toshow that σnk − (U ′σ∩ σnk ) is closed in σnk ∀ k, which is equivalent under topology of cells for U ′σ∩ σnk to beopen in σnk . For arbitrary k we have U ′σ ∩ σnk = (Uσ ∩ σnk )

⋃j(fj,n(Wj,σ)∩ σnk ) = (Uσ ∩ σnk )

⋃fk,n(Wk,σ)

and taking the preimage we have Y = f−1k,n(U ′σ ∩ σnk ) = f−1

k,n(Uσ) ∪Wk,σ. The n-disk is compact, theCW-complex Xn is Hausdorff, a closed subset of a compact space is compact (Theorem 26.2[6]), theimage of a compact set under a continuous map is compact (Theorem 26.5[6]), and every compact subsetof a Hausdorff space is closed (Theorem 26.3[6]); thus fk,n is a closed map and hence a quotient map forσnk (by Theorem 22.1[6]). It suffices to check that Y ⊂ Bnk is open, for then fk,n(Y ) = U ′σ ∩ σnk is openin σnk by definition of a quotient map. By construction, Y = x ∈ Bnk − bk | r(x) ∈ f−1

k,n(Uσ) where

r : Bnk − bk → ∂Bnk ≡ Sn−1k is the radial projection r(x) = x−bk

||x−bk|| . As r is continuous and f−1k,n(Uσ) is

open in ∂Bnk , we have Y = r−1(f−1k,n(Uσ)) open in Bnk − bk and hence in Bnk (by Lemma 16.2[6]). Our

new collection for Xn is the open sets U ′σ (where σ ⊂ Xn−1) plus the open n-cells U ′σnj = σnj ; this is the

canonical open cover of Xn and hence completes the induction.Any point x ∈ X will lie in an i-cell σ which lies in the open set U ′σ, and we take this as our desiredneighborhood of x: since any open set of our constructed cover is bounded by barycenters, and g ∈ Gmaps barycenters to barycenters by construction, we have gU ′σ = U ′gσ which is disjoint from U ′σ byconstruction for all g 6= 1.

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4.3: Given G = Z ⊕ Z we have the torus T with π1T ∼= G and its universal cover ρ : R2 → T .After drawing the lattice Z2 ⊂ R2, pick a square and label that surface L, with the bottom left cor-ner as the basepoint x0 and the bottom side as the edge es and the left side as the edge et (so thecorners are x0, sx0, stx0, tx0 going counterclockwise around L, and the top and right sides of L are re-spectively tes and set). Following Brown’s notation [in this section], x0 generates C0(R2) and es, etgenerate C1(R2) with ∂1(es) = (s − 1)x0 and ∂1(et) = (t − 1)x0. Lastly, L generates C2(R2) with∂2(L) = es + set − tes − et = (1− t)es − (1− s)et. Thus the desired free resolution of Z over ZG is:

0→ ZG ∂2−→ ZG⊕ ZG ∂1−→ ZG ε−→ Z→ 0 .

4.4:

5.1: The homotopy operator h in terms of the Z-basis g[g1| · · · |gn] for Fn is h(g[g1| · · · |gn]) = [g|g1| · · · |gn].

5.2: Using G = Z2 = 1, s, the elements of the normalized bar resolution F∗ = F∗/D∗ are [s|s| · · · |s],and each element forms a basis for the corresponding dimension, giving the identification Fn = ZG.Denoting si = s ∀ i,

di[s1|s2| · · · |sn] =

s[s| · · · |s] , i = 0

[s1| · · · |si−1|sisi+1| · · · |sn] = 0 , 0 < i < n

[s| · · · |s] , i = n

The middle equation resulted from sisi+1 = s2 = 1 (so the element lies in D∗). The boundary operator∂n then becomes s− 1 for n odd and s+ 1 for n even.∴ the normalized bar resolution is:

· · · → ZG s−1−→ ZG s+1−→ ZG s−1−→ ZG ε→ Z→ 0 .

5.3(a): [[Geometric Realization of a Semi-Simplicial Complex ]]For each (n+1)-tuple σ = (g0, . . . , gn), let ∆σ be a copy of the standard n-simplex with vertices v0, . . . , vn.Let diσ = (g0, . . . , gi, . . . , gn) and let δi : ∆diσ → ∆σ be the linear embedding which sends v0, . . . , vn−1

to v0, . . . , vi, . . . , vn. Consider the disjoint union X0 =∐σ ∆σ (topologize it as a topological sum) and

define the quotient space Xdef= X0/∼ using the equivalence relation generated by (σ, δix) ∼ (diσ, x),

where we rewrite ∆σ as σ ×∆σ for clarity of the relation properties.We assert that the geometric realization X is a CW-complex with n-skeleton Xn = (

∐dim∆σ≤n ∆σ)/∼.

X0 is the collection of vertices and hence a 0-skeleton, and so we proceed by induction on n [sketch]:

In Xn the equivalence relation ∼ identifies a point on a boundary ∂∆(n)σ with a point in Xn−1, and

it doesn’t touch the interior points of ∆(n)σ . This means that the n-cells are ∆(n)

σ with the attachingmaps induced by di ∀ i. Refer to Theorem 38.2[4] for the analogous construction with adjunction spaces,providing Hausdorffness of Xn and weak topology w.r.t. Xii<n. Thus, X is a CW-complex as theunion

⋃Xi with the weak topology.

We define the G-action on the simplices by left multiplication on their associated tuples; it is free sincetuples are unique and so the only element which fixes a tuple (and hence a simplex) is the identityelement of G. This makes X a G-complex. We deduce that X is contractible because for each simplex,X contains its cone which is contractible, so taking hσ = (1, g0, . . . , gn) we can use the straight-linehomotopy between δ0 : ∆σ → ∆hσ and the constant map ∆σ → ∆hσ at v0 (for any point in the domainsimplex of this homotopy H which lies on a subsimplex, the homotopy associated to that subsimplex isjust the restriction of H, and hence the straight-line homotopy is well defined).To form the desired isomorphism between the cellular chain complex C(X) and the standard resolutionF∗, it suffices to determine the boundary operator on C(X) and see that it provides commutativity ofthe diagram F∗ → C(X), noting that Ci(X) ∼= Fi by the correspondence ∆σ ↔ σ. Now C(X0) has theboundary ∂[v0, . . . , vn] =

∑i(−1)i[v0, . . . , vi, . . . , vn] and each resulting (n-1)-simplex is the image under

a δi. Thus C(X) has boundary maps ∂(∆σ) =∑i(−1)i∆diσ, and these parallel those of F∗, giving

commutativity of the diagram.

Note that this provides a solution to Exercise I.4.4: For any group G we can form the contractible

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G-complex X as above, and since X → X/G is a regular covering map by the result of exercise I.4.2above, the orbit space X/G is a K(G, 1), called the “classifying space.”

5.3(b): For the normalized standard resolution F∗, we simply follow part(a) while making these furtheridentifications inX0 to collapse degenerate simplices. For each σ = (g0, . . . , gn) let siσ = (g0, . . . , gi, gi, . . . , gn),and when forming the quotient X0 → X we also collapse ∆siσ to ∆σ via the linear map Li whichsends v0, . . . , vn+1 to v0, . . . , vi, vi, . . . , vn (so the only simplices of X are those whose associated tu-ples have pairwise distinct coordinates). Thus the equivalence relation in part(a) is also generated by(σ, Lix) ∼ (siσ, x). During the inductive process for Xn, if ∆σ is a degenerate simplex then Xn−1 willalready contain it and so those simplices need not be considered as n-cells. No problems arise when usingthe homotopy because for tuples of the form τ = (1, g0, . . . , gn) | g0 6= 1, the cone ∆τ ∗ v0 = ∆hτ = ∆τ

has the identity map δ0 still being nulhomotopic [note: we actually have a deformation retraction since∆(1) remains fixed instead of looping around ∆(1,1) as in part(a)].

6.1: Given a finite CW-complex X with a map f : X → X such that every open cell satisfiesf(σ) ⊆

⋃τ 6=σ τ where dimτ ≤ dimσ, we have the condition f(σ) ∩ σ = ∅ and there are no fixed

points. Viewing the open n-cell σ on the chain level in Hn(X(n), X(n−1)) = Cn(X), f](σ) does notconsist of σ and so the respective matrix has the value 0 at the row-column intersection for σ. Therefore,tr(f], Cn(X)) = 0 ∀ n since the diagonal of the matrix for the basis elements is zero. By the Hopf TraceTheorem,

∑(−1)itr(fi, Hi(X)/torsion) =

∑(−1)itr(f], Ci(X)) and so the Lefschetz number Λ(f) = 0.

If X is a homology (2n − 1)-sphere, then Hi(X) is nontrivial only in dimensions 0 and 2n − 1 (inwhich case it is isomorphic to Z). Thus, by the Lefschetz Fixed Point Theorem, (−1)0tr(f0,Z) +(−1)2n−1tr(f2n−1,Z) = 1− d = 0 ⇒ d = 1 and f∗ : H2n−1(X)→ H2n−1(X) is the identity.

6.2: The group action G→ Homeo(S2n) yields a degree map

φ : G→ Aut(H2n(S2n)) ∼= Aut(Z) = ±1 = Z/2Z

which sends g ∈ G to the degree d = deg(g) of its associated homeomorphism g : S2n → S2n [note:deg(g) · deg(g−1) = deg(g · g−1) = deg(id) = 1 ⇒ |d| = 1].Consider nontrivial G 6= Z/2Z and assert that this φ is not injective:Kerφ = 0 ⇒ 3 ≤ |G| = |Kerφ| · |Imφ| = 1 · |Imφ|. Since Imφ ⊆ Z/2Z, |Imφ| = 1 or 2 . In either casewe arrive at a contradiction (since 1, 2 < 3). ∴ ∃ g ∈ Kerφ | g 6= id ⇒ degg = 1. Now assume thisaction is free, and use the notation fi : Hi(S

2n) → Hi(S2n). By the Lefschetz Fixed Point Theorem,

(−1)0tr(f0) + (−1)2ntr(f2n) = 1 + d = 0 ⇒ degf = d = −1 ∀ f 6= id. Our contradiction has now beenreached (taking f = g from above).

7.1: Given the finite cyclic group G = 〈t〉, the free resolution F of Z over ZG with period two (chaincomplex with rotations of S1), and the bar resolution F ′, we obtain a commutative diagram wheref : F → F ′ is the desired augmentation-preserving chain map:

· · · // ZG

f3

t−1// ZG

f2

N // ZG

f1

t−1// ZG

f0

ε // Z

idZ

// 0

· · · // F ′3 // F ′2 // F ′1 // F ′0 // Z // 0

We define f inductively as fn+1 = knfn∂, where k is a contracting homotopy for the augmented complexassociated to F ′, and each map is determined by where it sends the basis element:f0(1) = k−1idZε(1) = k−1idZ(1) = k−1(1) = (1) ≡ [ ]f1(1) = k0f0∂(1) = k0f0(t− 1) = k0tf0(1)− f0(1) = k0t[ ]− [ ] = [t]− [1]

f2(1) = k1f1∂(1) = k1f1(N) = k1∑n−1i tif1(1) =

∑n−1i ([ti, t]− [ti, 1])

f3(1) = k2f2∂(1) = k2f2(t− 1) =∑n−1i ([t, ti, t] + [1, ti, 1]− [t, ti, 1]− [1, ti, t])

7.2: Here is the axiomatized version, Lemma 7.4:Under the additive category A, let (C, ∂) and (C ′, ∂′) be chain complexes, let r be an integer, and let(fi : Ci → C ′i)i≤r be a class of morphisms such that ∂′ifi = fi−1∂i for i ≤ r. If Ci is projective relativeto the class E of exact sequences for i > r, and C ′i+1 → C ′i → C ′i−1 is in E for i ≥ r, then (fi)i≤r extendsto a chain map f : C → C ′ and f is unique up to homotopy. (Theorem 7.5 follows immediately)

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7.3(a): Given an arbitrary category C, let A ∈ Ob(C) be an object and let hA = HomC(A,−) : C →(Sets) be the covariant functor represented by A, with uA ∈ hA(A) as the identity map A → A. LetT : C → (Sets) be an arbitrary covariant functor. Any natural transformation ϕ : hA → T yields thecommutative diagram (with f : A→ B in hA(B) arbitrary):

hA(A)hA(f)−−−−→ hA(B)yϕ yϕ

T (A)T (f)−−−−→ T (B)

For any v ∈ T (A) suppose we have the natural transformation with ϕ(uA) = v. By commutativity,T (f)(v) = (ϕ hA(f))(uA) = ϕ(f uA) = ϕ(f), and hence the transformation is unique [determined bywhere it sends the identity]. For existence of the natural transformation ϕ(f) = T (f)(v), we assert thatit satisfies the commutative diagram, using arbitrary g : B → C (and f as above):

T (g)[ϕ(f)] = T (g)[T (f)(v)] = T (g f)(v) = ϕ(g f)ϕ[hA(g)(f)] = ϕ(g f) = T (g)[ϕ(f)]

Thus, HomF(hA, T ) ∼= T (A) where F is the category of functors C→ (Sets), and we have finished provingYoneda’s Lemma.

7.3(b): An M-free functor F : C → Ab is isomorphic to⊕

α ZhAα , where Aα ∈ M [M is a sub-class of Ob(C)] and ZhA : C→ Ab is the composite of hA and the functor (Sets)→ Ab which associatesto a set the free abelian group it generates. Given the additive category A whose objects are covariantfunctors C→ Ab and whose maps are natural transformations of functors, let E be the class of M-exactsequences in A. Consider the mapping problem (for all rows in E):

Fψ

~~

ϕ

0

T ′i// T

j// T ′′

By Yoneda’s Lemma (part(a) above), each component ZhAα of F with any natural transformation in theabove diagram is completely determined by the identity uAα , and so these identities “form a basis” forF . In particular, for the identity uA we obtain the exact sequence T ′(A) → T (A) → T ′′(A) of abeliangroups from the above mapping problem since the associated row lies in E with A ∈M. Thus, for eachidentity we have ϕ(uAα) ∈ Kerj = Imi, which implies ∃ xα ∈ T ′(Aα) | i(xα) = ϕ(uAα), and so weform ψ by ψ(uAα) = xα. This means that F (an M-free functor) is projective relative to the class E ofM-exact sequences.

7.3(c): There is a natural chain map in A from M-free complexes to M-acyclic complexes, and itis unique up to homotopy [using the categorial definitions from parts (a) and (b)]. This statement is aresult of the combination of part(b) and Exercise 7.2 above, and is precisely the Acyclic Model Theoremin a rephrased form (M is the set of models).

7.4: Under the category of R-modules, let (C, δ) and (C, δ) be cochain complexes, let r be an inte-ger, and let (fi : Ci → Ci)i≤r be a class of morphisms such that fi−1δi−1 = δi−1f

i for i ≤ r. If Ci is

injective relative to the class E of exact sequences for i > r [yielding a cochain complex of injectives],and Ci−1 → Ci → Ci+1 is in E for i ≥ r [an acyclic cochain complex], then (fi )i≤r extends to a cochainmap f : C → C and f is unique up to homotopy.(The analogous “Theorem 7.5” follows immediately)

8.1: Obviously the trivial group is one, since Z[1] = Z and Z is a projective Z-module; so assume G isnontrivial. Give Z the trivial module structure (so that for r ∈ ZG, r · a = ε(r)a ∀ a ∈ Z). Considering

the short exact sequence of modules 0→ I → ZG ε→ Z→ 0, we must find a splitting µ : Z→ ZG for Zto possibly be ZG-projective. Any such map is determined by where 1 ∈ Z is sent; say µ(1) = x. Then,for nontrivial α =

∑rigi, µ(α ·1) = α ·µ(1) = α ·x = αx. But µ(α ·1) = µ(

∑ri) = µ(1)

∑ri = (

∑ri)x,

so αx = (∑ri)x. Thus (α −

∑ri)x = 0 ⇒

∑rigi =

∑ri ⇒

∑ri(gi − 1) = 0 [this sum can be viewed

9

as having no gi = 1, and hence it lies in I]. Restricting our choice of nontrivial α to one which is not aninteger, there is some nontrivial ri0 associated to gi0 6= 1 and hence we must have gi = 1 ∀ i (by freenessof I). But then G is the trivial group, and we are done.

8.2: Assume P is a projective ZG-module and consider the subgroup H ⊆ G. Then F = P ⊕ Kwhere F is a free ZG-module, by Proposition I.8.2[1]. By restriction of scalars from ZG to its subringZH [r · n = f(r)n with homomorphism f : ZG → ZH preserving identities], F has an inherent ZH-module structure. Since P is a direct summand of such a free module, it is a projective ZH-module, byProposition I.8.2[1]. Alternatively, we can note from Exercise 3.1 above that ZG =

⊕ZH and so F is a

direct sum of H-modules, hence ZH-free.

8.3(a): Given F as a non-negative acyclic chain complex of projective modules Pn (over an arbitrary

ring R), it will be contractible if each short exact sequence 0→ Zn → Pn∂n→ Zn−1 → 0 splits (where ∂ is

induced by the boundary ∂), by Proposition I.0.3[1]. The case n = 0 is trivial using the zero map, andso arguing inductively we assume that ∂n−1 splits. Since Ker∂n−1 = Zn−1 is a direct summand of theprojective module Pn−1 and a projective module is a direct summand of a free module by PropositionI.8.2[1], Zn−1 is necessarily a direct summand of a free module, and hence is projective by Lemma I.7.2[1].Therefore, ∂n must have a splitting in the aforementioned short exact sequence by Proposition I.8.2[1].

8.3(b): If R is a principal ideal domain, then submodules of a free R-module are free by TheoremII.7.1[5]. Since a projective module is a direct summand of a free module by Proposition I.8.2[1], it isin particular a submodule and hence is free (over R as a PID). Therefore, submodules of a projectivemodule over a PID are free and hence projective (free modules are projective by Lemma I.7.2[1]). Thenon-negativity hypothesis of Corollary I.7.7 can be dropped if we then restrict ourselves to PIDs, becausewe can follow part(a) above but not use induction since Zn−1 is already projective, being a submoduleof the projective chain module Pn−1 (i.e. we don’t need any “starting point” in the resolution to obtainthe desired splitting).

8.4: Every permutation module admits the decomposition QX ∼=⊕Q[G/Gx] and a direct sum of

projective modules is projective iff each summand is projective (by Lemma XVI.3.6[5]). Thus it sufficesto show that Q[G/Gx] is a projective QG-module, where G is an arbitrary group and Gx is finite. Notethat HomQG(Q[G/Gx],−) is a left-exact functor (Corollary 10.5.32[2]); it is given by M →MGx becauseany homomorphism ϕ is determined by ϕ(Gx), and ϕ(Gx) = ϕ(g · Gx) = g · ϕ(Gx) for any g ∈ Gx.Thus, we must show that HomQG(Q[G/Gx],−) takes surjective homomorphisms M → M to surjectivehomomorphisms MGx → MGx [because then the functor is exact and then Q[G/Gx] is projective bydefinition]. If m ∈ MGx , lift m to m ∈M . Then 1

|Gx|∑g∈Gx gm is also a lifting of m which lies in MGx

(because 1|Gx|

∑gm 7→ 1

|Gx|∑gm = 1

|Gx| · |Gx|m = m), and so MGx → MGx is surjective. Thus, QX is

a projective QG-module, where X is a G-set and Gx is finite for all x ∈ X.

8.5: If G is finite and k is a field of characteristic zero, consider any short exact sequence of kG-

modules of the form 0 → M ′ → Mϕ→ P → 0. Since k-vector spaces are free modules over k, and free

modules are projective (by Lemma I.7.2[1]), P is k-projective and hence the sequence [as k-modules]splits by Proposition I.8.2[1]. Choosing a splitting f : P → M for the underlying sequence of k-vectorspaces, we form the homomorphism φ : P →M by x 7→ 1

|G|∑g∈G gf(g−1x).

Since f is a k-module homomorphism, it suffices to show that φ is equivariant (compatible with G-action)for it to thus be a kG-module homomorphism:

φ(g0x) = 1|G|∑gf(g−1g0x) = 1

|G|∑g0hf(h−1x)

[where h = g−10 g]

g0φ(x) = g0( 1|G|∑gf(g−1g0x)) = g0( 1

|G|∑hf(h−1x)) = φ(g0x)

[because∑g∈G h =

∑g∈G g, as g−1

0 permutes the elements of G]

By Proposition 10.5.25[2] it suffices to show that ϕφ = idP for φ to thus be a splitting (noting thatϕf = idP ):

ϕ(φ(x)) = ϕ( 1|G|∑g∈G gf(g−1x)) = 1

|G|∑g∈G ϕ(gf(g−1x)) = 1

|G|∑g∈G gϕ(f(g−1x)) =

10

1|G|∑g∈G gg

−1x = 1|G|∑g∈G x = 1

|G| |G|x = x

Since φ is a kG-splitting, P is kG-projective by Proposition I.8.2[1].

8.6: Suppose P is an R-module such that ϕ : P ∗ ⊗R P → HomR(P, P ) is surjective, where thismap is given by ϕ(u ⊗ m)(x) = u(x) · m. Then in particular we have idP = ϕ(

∑fi ⊗ ei), so that

x = idP (x) =∑ϕ(fi ⊗ ei)(x) =

∑fi(x)ei. Thus, P is projective by Proposition I.8.2[1], and it is

finitely generated by the ei elements (noting that the summation is finite for tensor products).

8.7: Assume P is a finitely generated projective R-module and M is any (left) R-module, and takethe canonical isomorphism ϕ : P ∗ ⊗R M → HomR(P,M) from Proposition I.8.3[1] which is given byϕ(u ⊗m)(x) = u(x) ·m. For any z ∈ P ∗ ⊗R P we define the map ψz : HomR(P,M) → P ∗ ⊗R M asψz(f) = (P ∗ ⊗ f)(z), which is a homomorphism since tensors are distributive over sums.

Method 1 : View ϕ−1 as a natural transformation HomR(P,−) → P ∗ ⊗R −, where HomR(P,−)and P ∗ ⊗R − are exact covariant functors from the category R-mod to the category Ab by Corollary10.5.41[2] and Corollary 10.5.32[2], noting that P ∗ is projective by Proposition I.8.3[1] and hence is aflat module by Corollary 10.5.42[2]. By Yoneda’s Lemma, ϕ−1 is uniquely determined by ϕ−1(idP ) = z,and the proof of the lemma (Exercise I.7.3(a) above) states that ϕ−1(f) = (P ∗ ⊗ f)(z) = ψz(f). Thus,the inverse homomorphism ϕ−1 is a map of the form ψz (independent of M).

Method 2 : By Proposition I.8.2[1] we can choose elements ei ∈ P and fi ∈ P ∗ such that for everyx ∈ P , x =

∑fi(x)ei and fi(x) = 0 for cofinitely many x. Set z =

∑fi ⊗ ei. We have ψzϕ = id

because ψz(ϕ(u ⊗ m)) = (P ∗ ⊗ u · m)(z) =∑fi ⊗ u(ei) · m =

∑fi · u(ei) ⊗ m = u ⊗ m [note:

u(x) = u(∑fi(x)ei) =

∑fi(x) · u(ei) as u is an R-module homomorphism]; we also have ϕψz = id

because ϕ(ψz(f)) = ϕ[(P ∗ ⊗ f)(z)] = ϕ(∑fi ⊗ f(ei)) =

∑fi · f(ei) = f [note: f(x) = f(

∑fi(x)ei) =∑

fi(x) · f(ei) as f is also an R-module homomorphism]. Thus, the inverse ϕ−1 is a map of the form ψz(independent of M).

8.8: Since P is finitely presented, we can form the obvious exact sequence F1 → F0 → P → 0 with F0 andF1 free of finite rank (the generators and relators, respectively). By Theorem 10.5.33[2], HomR(−, D)is a left exact contravariant functor, and so we obtain an exact sequence 0 → P ∗ → F ∗0 → F ∗1 of rightR-modules [notation: M∗ = HomR(M,R)]. Since P is a flat module, we can tensor this exact sequencewith P to obtain the exact sequence 0 → P ∗ ⊗R P → F ∗0 ⊗R P → F ∗1 ⊗R P . Since Fi [i = 0, 1] isfree, it is necessarily projective (by Lemma I.7.2[1]) and hence F ∗i ⊗R P ∼= HomR(Fi, P ) by PropositionI.8.3[1]. Since F1 is contained in the quotient of F0 which gives P , HomR(F1, P ) = 0 and thus we havethe isomorphism P ∗ ⊗R P ∼= HomR(F0, P ).Now HomR(−, P ) as a functor on the original presentation sequence gives rise to the exact sequence 0→HomR(P, P ) → HomR(F0, P ) → HomR(F1, P ) = 0, and thus we have the isomorphism HomR(P, P ) ∼=HomR(F0, P ).

Since the discovered isomorphism P ∗ ⊗R P∼=→ HomR(P, P ) is surjective, P is a projective R-module by

Exercise I.8.6 above.

11

2 Chapter II: The Homology of a Group

2.1: For S an arbitrary G-set, ZS ∼=⊕Z[G/Gs]. By the Orbit-Stabilizer Theorem we have a bijection

between Gs and G/Gs. When passing to the quotient for the group of co-invariants, the subset Gs ⊆ Sis sent to the element Gs ⊆ S/G since s is identified with gs in S/G (giving trivial G-action). Therefore(ZS)G ∼= (

⊕Z[G/Gs])G ∼=

⊕(Z[G/Gs])G ∼=

⊕Z[Gs] ∼= Z[S/G].

2.2: A weaker hypothesis “Let X be an arbitrary G-complex without inversions” suffices. Then C∗(X) ∼=⊕Z[Xi] is a direct sum of permutation modules where Xi is the basis set of i-cells, so by the previous

exercise, C∗(X)G ∼=⊕

(Z[Xi])G ∼=⊕Z[Xi/G] ∼= C∗(X/G) which has a Z-basis with one basis element

for each G-orbit of cells of X.

2.3(a): With the G-module M and normal subgroup H /G, MH = M/IM where I is the augmentationideal of ZH. The induced G/H-action is given by gH(m+IM) = gm+IM . The properties of an actionare obviously satisfied, and it is well defined because if g2 is another coset representative of g1H, theng2 = g1h and g2m+IM = g1hm+IM = g1hm−g1m+g1m+IM = (h−1)g1m+g1m+IM = g1m+IM .

2.3(b): We can form the group homomorphism ϕ : MG → (MH)G/H using part(a) by m 7→ m+ IM ,

which is well-defined because m = gm 7→ gm+ IM = gH(m+ IM) = m+ IM ; it is a homomor-phism since ϕ(m1 m2) = ϕ(m1m2) = m1m2 + IM = (m1 + IM)(m2 + IM) = (m1 + IM)(m2 + IM) =ϕ(m1)ϕ(m2). For the inverse φ we use m+ IM 7→ m which is well-defined since given the equivalentelements m+ IM and gm+ (h− 1)m′ + IM in (MH)G/H ,

φ(gm+ (h− 1)m′ + IM) = gm+hm′−1m′ = m+m′−m′ = m = φ(m+ IM); it is a homomorphism be-cause φ(m1 + IM m2 + IM) = φ(m1m2 + IM) = m1m2 = m1 m2 = φ(m1 + IM)φ(m2 + IM). Thus,we have the isomorphism MG

∼= (MH)G/H .

2.3(c): Let Z[G/H]⊗ZGM be a G/H-module, where Z[G/H] is the obvious (G/H,G)-bimodule whichforms the tensor product and gives it the desired module structure. The map Z[G/H]×M →MH givenby (a,m) 7→ am is clearly G-balanced, and so by the universal property of tensor products (Theorem10.4.10[2]) there exists the group homomorpism ϕ : Z[G/H]⊗ZGM → MH given by a⊗m 7→ am, andit is clearly a G/H-module homomorphism. There is a well-defined map φ : MH → Z[G/H] ⊗ZG Mdefined as m 7→ 1⊗m because of the identity 1H ⊗ hm = 1Hh⊗m = 1H ⊗m, and it is a G/H-modulehomomorphism because φ(gH · m) = φ(gm) = 1H⊗gm = Hg⊗m = gH⊗m = gH1H⊗m = gH ·φ(m)[noting that gH = Hg since H is normal, and G/H acts on MH by part(a) above]. Since ϕ and φ areinverses of each other, they are isomorphisms and we obtain MH

∼= Z[G/H]⊗ZGM .

3.1: Let g1, . . . gn ∈ G be pairwise-commutative elements and consider z =∑

(−1)sgnσ[gσ(1)| · · · |gσ(n)] ∈Cn(G), where σ ranges over all permutations of 1, . . . , n. The sign of a permutation is defined here tobe the number of swaps between adjacent integers to bring the permuted set back to the identity. Lookingat the boundary ∂z where ∂ =

∑(−1)idi, a particular dj with j 6= 0, n will provide elements in Cn−1(G)

of the form [· · · |gkgk′ | · · · ]. Each of these appears twice because gkgk′ = gk′gk, but the paired elementswill have opposite signs and hence will cancel each other. For j = 0, n we have elements of the form[gi| · · · ] with one gk missing from each, and each of these elements also appears twice because d0 willtake off gk from the beginning of some element while dn will take off gk from the end of some otherelement. The paired elements differ by the sign (−1)n due to the boundary map, and they also differby the sign (−1)n−1 due to the permutation which takes the first slot and sends it to the last slot; since(−1)n(−1)n−1 = (−1)2n−1 = −1, these paired elements will also cancel each other. Thus, ∂z = 0 and zis a cycle in Cn(G).

3.2: Suppose Z admits a projective resolution of finite length over ZG where G = Zn. Then ∃ i0 | HiG =0 ∀ i > i0, and we make note that HiG is independent of the choice of resolution (see Section II.1[1]).Yet by an earlier calculation II.3.1[1] (using an infinite resolution), HiG ∼= Zn for all positive odd integersi. Thus we have arrived at a contradiction.

3.3: If G has torsion, say Zn ⊆ G, and Z admits a projective resolution of finite length over ZG,

12

then since a projective ZG-module is projective as a ZH-module for any subgroup H ⊆ G (by ExerciseI.8.2), we would obtain a corresponding finite projective resolution over Z[Zn]. But this cannot occurdue to the previous exercise, and hence we arrive at a contradiction.

4.1: If Y is a path-connected space and has a contractible regular covering space X with coveringgroup G, then X is its universal cover with free G-action as translation, and π1Y = G (so Y ∼= X/G isa K(G, 1)-space). The singular chain module Csingn (X) is a free Z-module with basis the set of singularsimplices σn : ∆n → X (continuous maps of the standard simplex into the space). A G-action on thisbasis is given by the composition gσn : ∆n → X → X, and thus Csingn (X) is a free ZG-module withone basis element for every G-orbit of singular simplices. Denoting the ith face of σn as σnFi whereFi = [v0, . . . , vi, . . . , vn] : ∆n−1 → ∆n is the inclusion map, the induced G-action on the simplices mapsfaces of σn to faces of gσn. Thus, the boundary operator is equivariant and the singular [augmented]chain complex Csing∗ (X) is a free G-module chain complex. As X is contractible, the complex is exactand so it is a free resolution of Z over ZG.Consider the projection ϕ : Csing∗ (X) → Csing∗ (Y ) given by σ 7→ qσ, where q : X → X/G ∼= Y isthe regular covering map. Noting that ϕ(gσ) = qgσ = qσ = ϕ(σ), the projection induces the mapϕ : Csing∗ (X)G → Csing∗ (Y ) which sends basis elements to basis elements. Since Csing∗ (Y ) has a Z-basiswith one basis element for each G-orbit of singular cells of X as does Csing∗ (X)G [by a property (II.2.3)of the coinvariants functor], ϕ is an isomorphism. Therefore, H∗G ∼= H∗Y as the homologies of a groupare independent of the choice of resolution up to canonical isomorphism.

4.2:

4.3:

5.1: Let Y be an n-dimensional connected CW-complex such that πiY = 0 for i < n (n ≥ 2), let π = π1Y ,and let X be the universal cover of Y (so that π1X = 0). Since πiX ∼= πiY for i > 1, πiX is trivial fori < n and so by the Hurewicz Theorem HiX = 0 for 0 < i < n and the Hurewicz map h : πnX → HnXis an isomorphism. In addition, we have a partial free resolution Cn(X)→ · · · → C0(X)→ Z→ 0 whosenth homology group is ZnX = HnX (noting that X is n-dimensional). Lemma II.5.1[1] now gives us anexact sequence 0 → Hn+1π → (HnX)π → ZnY = HnY → Hnπ → 0, where Y ∼= X/π because everyuniversal cover is regular with covering transformation group π1Y (by Corollary 81.4[6]). Finally, notingthat the coinvariants functor takes isomorphisms to isomorphisms (by right-exactness), the commutative

diagram (πnX)π

∼=

∼= // (HnX)π

(πnY )π // HnY

yields the desired exact sequence 0→ Hn+1π → (πnY )π → HnY → Hnπ → 0.

5.2(a): Let G = 〈S ; r1, r2, . . .〉 = F (S)/R where R is the normal closure in F (S) of the words ri.Consider the abelianization map rirj 7→ [ri]+[rj ] from R to the relation module Rab with the denotation[ri] = rimod[R,R]. Any element x ∈ R has a representation x =

∏ni=1(fir

±1i f−1

i ) where fi ∈ F = F (S).Now F acts by conjugation on R and so induces an F -action on Rab, and R acts trivially on Rabdue to the definition of abelianization; it is immediate that we obtain the (G = F/R)-action on Rab:g · [ri] = [frif

−1]. Subsequently, [x] =∑ni=1(gi · [ri]) and hence Rab is generated as a G-module by the

images of the presentation words.

5.2(b): Let Y = (∨s S

1) ∪r1 e2 ∪r2 e2 ∪ · · · be the 2-complex associated to the given presentation

of G in part(a), and let Y be its universal cover. Consider the augmented cellular chain complex

(∗) 0→ C2(Y )∂2→ C1(Y )

∂1→ C0(Y )→ Z→ 0

Note that C0(Y ) = ZG and C1(Y ) =⊕|S| ZG and C2(Y ) =

⊕|R| ZG. By Proposition II.5.4[1] we have

the exact sequence

0 // Rab // C1(Y )∂1 // C0(Y ) // Z // 0

13

and hence Rab = Ker∂1 is always in the chain complex for Y . If Y is a K(G, 1) then Y is acyclic and(*) is exact by Proposition I.4.2[1], so C2(Y ) = Rab and hence Rab is a free ZG-module.Now suppose Y is not acyclic, so that Y is not a K(G, 1). Then H2Y is nontrivial (since HiY = 0 fori > 2 by (*) and H1Y = 0 by the simply-connected property of Y ) which implies that the boundary map∂2 is not injective, and by exactness we can refer to this non-injective map as C2(Y ) → Ker∂1 = Rab.Therefore, there exists a nontrivial ZG-relation amongst the words ri in Ker∂2, so [ri] is not ZG-independent in Rab and hence does not generate Rab freely.

5.2(c): Let G = 〈S; r〉 be an arbitrary one-relator group and write r = un ∈ F = F (S), wheren ≥ 1 is maximal. By a result of Lyndon-Schupp, the image t of u in G has order exactly n, and we letC = 〈t〉 = Zn. If n > 1 then the relation module Rab is not freely generated by rmod[R,R] since thisgenerator is fixed by C, but a result of Lyndon shows that no other relations hold (i.e. the projectionZ[G/C]→ Rab is an isomorphism).Let Y be a bouquet of circles indexed by S, and let Y be the connected regular covering space of Ycorresponding to the normal subgroup R of F = π1Y . Choosing a basepoint v ∈ Y lying over thevertex of Y , we identify G with the group of covering transformations of Y ; as explained in [1] onpg15, Y is a (1-dimensional) free G-complex. Since u ends at tv, the lifting r is the composite path

tv

tu

vuoo

t2v // tn−1v

tn−1u

OO

Thus the map S1 → Y corresponding to r is compatible with the action of C, where C acts on S1 as agroup of rotations (i.e. tk is multiplication by e2πik/n). Consider the 2-complex X obtained by attaching2-cells to Y along the loops gr, where g ranges over a set of representatives for the cosets G/C. Givena 2-cell σ, each g ∈ G sends ∂σ homeomorphically to ∂σ′ for some 2-cell σ′, and so we just pick ourfavorite extension g : σ → σ′. This G-action makes X a G-complex since the permutations of 2-cells aredetermined by the permutations of their boundary loops. Thus, if σ is the 2-cell attached along r thenonly C ⊂ G will fix σ, since ti simply rotates the loop ∂σ (i.e. Gσ = C). Let Γ = (

∨s S

1) ∪r e2 be the

standard 2-complex associated to the presentation of G; its universal cover is Γ = Y⋃g∈G σg where σg

is attached along gr, and C2(Γ) = ZG. Thus X is the quotient of Γ by identifying σg with σgti for all i

(for each g), and C2(X) = Z[G/C] ∼= Rab. If n = 1 then C = 1 and X = Γ. Lyndon’s theorem aboutone-relator groups says that Rab is freely generated by the image of r, provided r is not a power, whichis equivalent to Γ being a K(G, 1) by part(b) above, and hence equivalent to X being contractible (X isthe Cayley complex associated to the presentation of G).

5.3(a): With G = F/R and following Kenneth Brown’s proof of Theorem II.5.3, let F = F (S), letY be a bouquet of circles indexed by S, and let Y be the connected regular covering space of Y cor-responding to the normal subgroup R of F = π1Y . Choosing a basepoint v ∈ Y lying over the vertexof Y , we identify G with the group of covering transformations of Y . For any f ∈ F we regard f asa combinatorial path in the CW-complex Y and we denote by f the lifting of f to Y starting at v.This path f ends at the vertex f v, where f is the image of f in G. Define the function d : F → C1Yby letting df be the sum of the oriented 1-cells which occur in f . Since the lifting of f1f2 is the path

vf1−→ f1v

f1f2−→ f1f2v, we have d(f1f2) = df1 + f1df2 for all f1, f2 ∈ F . Thus, if we regard the G-moduleC1Y as an F -module via the canonical homomorphism q : F → G, then d is a derivation [since theF -action is given by restriction of scalars: f1 · df2 = f1df2 = q(f1)df2].

5.3(b): For any free group F we can apply part(a) above with R = 1 to get the desired deriva-tion d : F → Ω where G = F/1 = F and Ω = C1Y = ZF (S) which is the free module with basis (ds)s∈S[note: ds−1 = −s−1ds].The above note is a result of d(1) = d(1 · 1) = d(1) + 1d(1) = 2d(1) ⇒ d(1) = 0.We write the total free derivative df of f as the sum df =

∑s∈S(∂f/∂s)ds, where ∂f/∂s ∈ ZF is the

partial derivative of f with respect to s [the coefficient of ds when df is expressed in terms of the basis(ds)].It is immediate that ∂/∂s : F → ZF is a derivation because d is a derivation: ff ′ 7→ d(ff ′) =

14

df + fdf ′ =∑

[(∂f/∂s) + f(∂f ′/∂s)]ds 7→ (∂f/∂s) + f(∂f ′/∂s). For t ∈ S we have dt =∑

(∂t/∂s)ds =∑s6=t 0ds+ 1dt, and so ∂t/∂s = δs,t.

Example: S = s, t ⇒ ∂(ts−1ts2)/∂s = ∂ts−1/∂s+ ts−1(∂ts2/∂s) =[∂t/∂s+ t(∂s−1/∂s)] + ts−1[∂t/∂s+ t(∂ss/∂s)] =

0 + t[−s−1(∂s/∂s)] + ts−10 + ts−1t[∂s/∂s+ s(∂s/∂s)] = −ts−11 + ts−1t[1 + s1] = ts−1ts+ ts−1t− ts−1

5.3(c): Consider any free group F = F (S) and derivation d : F →M where M is an F -module. By therepresentation f = s±1

1 · · · s±1n and the definition of a derivation d(gh) = dg+ gdh, we have the equation

df =∑s∈S wsds through trivial induction on n, where ws = ∂f/∂s by definition of the partial derivative.

5.3(d): Consider θ in the exact sequence 0 → Rabθ→ ZG(S) ∂→ ZG ε→ Z → 0 of G-modules, where

ZG(S) is free with basis (es)s∈S and ∂es = s− 1 [bar denotes image in G], and consider ϕ : R→ ZG(S)

given by r 7→∑s∈S (∂r/∂s)es where (∂r/∂s) is the image of ∂r/∂s under the canonical map ZF → ZG.

In order to show that θ is induced by ϕ we must verify exactness of the above sequence, and so we startby calculating the partial derivatives of the representation r = sb11 · · · sbnn ∈ Rab (where si 6= sj 6=i). Since

∂sb/∂s =∑b−1j=0 s

j for b > 0 and ∂sb/∂s = −∑bj=1 s

−j for b < 0 and ∂(Asb)/∂s = A(∂sb/∂s) with s - A,we obtain

∂r/∂si =

sb11 · · · s

bi−1

i−1

∑bi−1j=0 sji bi > 0

−sb11 · · · sbi−1

i−1

∑bij=1 s

−ji bi < 0

Injectivity of ϕ|Rab follows immediately from the freeness of ZG(S) and the fact that any nontrivial r

has some nontrivial bi (hence ∂r/∂si 6= 0). It suffices to show that ∂ϕ(r) =∑

(∂r/∂s)(s− 1) = 0 for

r ∈ Rab. For a particular i, (∂r/∂si)(si−1) = s±b11 · · · s±bi−1

i−1 (∑s1±ji −

∑s±ji ) = s±b11 · · · s±bi−1

i−1 (s±bii −1),

and (∂r/∂si)(si − 1) + (∂r/∂si+1)(si+1 − 1) = −sb11 · · · sbi−1

i−1 + sb11 · · · sbii s

bi+1

i+1 [suppressing the ±]. Thus,∑(∂r/∂s)(s− 1) = −1 + 0 + · · ·+ 0 + sb11 · · · sbnn = −1 + r = −1 + 1 = 0, and exactness is satisfied.

If R is the normal closure of a subset T ⊆ F , then the projection ZG(T ) → Rab given by g · et 7→ g · [t]and the above exact sequence provides us with a partial free resolution:

ZG(T ) ∂2 //

## ##

ZG(S) ∂1 // ZG ε // Z // 0

Rab- θ

;;

where the matrix of ∂2 is the “Jacobian matrix” (∂t/∂s)t∈T,s∈S .

5.4: Sketch (via Ken Brown): Let G = F/R and use the same notation as in Exercise 5.3(a) above.Consider the following chain map in dimensions ≤ 2 using the bar resolution B∗

B2

Γ2

∂2 // B1

Γ1

∂1 // B0

Γ0

ε // Z // 0

ZG(R) // C1(Y ) // C0(Y ) // Z // 0

We have the identification C1(Y ) = ZG(S) = IR where I is the augmentation ideal of ZF (so IR is afree G-module on the images s− 1), and ZG(R) is the free G-module with basis (er)r∈R which mapsonto Rab = H1Y ⊂ C1(Y ). The specific chain map is given by Γ2 : [g1|g2] 7→ −g1g2 · r(g1, g2) andΓ1 : [g] 7→ f(g)− 1 and Γ0 : [ ] 7→ 1, where r(g1, g2) ∈ R and f(g) ∈ F such that f(g) = g ∈ G andf(g)f(h) = f(gh)r(g, h). Applying the coinvariants functor, the group homomorphism C2(G) → Rabgiven by [g|h] 7→ r(g, h) mod[R,R] induces the isomorphism ϕ : H2G → R ∩ [F, F ]/[F,R] by passage tosubquotients.A specific chain map γ in the other direction is given by γ2 : r 7→ hγ1(r − 1) and γ1 : s− 1 7→ [s] andγ0 : 1 7→ [ ], where h : B1 → B2 is the contracting homotopy h(g · [h]) = [g|h]. Regarding C2(G) as an

F -module via f · [g|h] = [fg|h], the map D : F → ZF → I → IRγ1→ B1

h→ B2 → (B2)G ≡ C2(G) is aderivation such that Ds = [1|s]. Moreover, Df =

∑s∈S [∂f/∂s|s], where the symbol [·|·] is Z-bilinear.

Then D|R : R → C2(G) is a homomorphism (since R acts trivially on C2(G)) which induces ϕ−1 bypassage to subquotients.

15

Let a1, . . . , an, b1, . . . , bn ∈ F such that r =∏ni=1[ai, bi] ∈ R. The formula ϕ−1(rmod[F,R]) =∑n

i=1[Ii−1|ai] + [Ii−1ai|bi] − [Ii−1aibia−1i |ai] − [Ii−1|bi], where Ii = [a1, b1] · · · [ai, bi], is proven in the

universal example where F is the free group on a1, . . . , an, b1, . . . , bn and R is the normal closure of r.Using the constructed formula for ϕ−1 and the product rule for derivations, the desired formula arisesfrom Dr =

∑i Ii−1 ·D[ai, bi] and D[a, b] = [1|a] + [a|b]− [aba−1|a]− [aba−1b−1|b].

5.5(a): With the presentation G = 〈s1, · · · , sn | r1, · · · , rm〉 we associate the 2-complex Y = (∨s S

1)∪r1e2 · · · ∪rm e2 so that π1Y ∼= G. By computing the Euler characteristic χ(Y ) two different ways (by The-orem 22.2[4]) we obtain the equation

∑(−1)irkZ(HiY ) =

∑(−1)ici, where rkZ is the rank and ci is the

number of i-cells. Then 1 − rkZ(Gab) + rkZ(H2Y ) = 1 − n + m, and so rkZ(H2Y ) = m − n + r wherer = rkZ(Gab) = dimQ(Q⊗Gab). Now H2Y = Ker∂2 is a free abelian group (subgroup of cellular 2-chaingroup), and by applying Theorem II.5.2[1] we get a surjection H2Y → H2G (from the exact sequence inthe theorem). Thus H2G can be generated by m− n+ r elements.

5.5(b): Since Gab = 0 and the number of generators equals the number of relations (in the finitepresentation), m − n + r = m − m + 0 = 0. Thus by part(a), H2G can be generated by at most 0elements, and so H2G = 0.

5.5(c): Given Gab = 0, H2G ∼= Z2 ⊕ Z2, and n as the number of generators, let m be the numberof relations in the presentation of G. Then H2G is generated by the two elements (0, 1) and (1, 0), andr = 0, so by part(a), 2 ≤ m− n+ 0⇒ m ≥ n+ 2. Therefore, any n-generator presentation must involveat least n+ 2 relations.

5.6(a): From the group extension 1→ N → G Q→ 1 we have G/N ∼= Q, and from Hopf’s formulawe have H2G ∼= R ∩ [F, F ]/[F,R] and H2Q ∼= S ∩ [F, F ]/[F, S], where G = F/R and Q = F/S withR ⊆ S ⊆ F (so N ∼= S/R). (H1N)Q ∼= (Nab)Q = Nab/〈(q−1) ·n[N,N ]〉 = Nab/〈gng−1n−1[N,N ]〉 =(N/[N,N ])/([G,N ]/[N,N ]) ∼= N/[G,N ], where the Q-action on Nab is induced by the conjugationaction of G on N , and the latter isomorphism follows from the Third Isomorphism Theorem. Now[G/N,G/N ] = g1Ng2Ng

−11 Ng−1

2 N = g1g2g−11 g−1

2 N = [G,G]N/N , so we have H1Q ∼= Qab =(G/N)/[G/N,G/N ] ∼= G/(N [G,G]), where the latter isomorphism follows from the Third IsomorphismTheorem.Thus, the desired 5-term exact sequence is obtained by showing the exactness of the sequence

R ∩ [F, F ]/[F,R]α→ S ∩ [F, F ]/[F, S]

β→ N/[G,N ]γ→ G/[G,G]

δ→ G/(N [G,G])→ 0

where γ and δ are induced by the injection and surjection of the group extension. From δ : g[G,G] 7→g[G,G]N = gN [G,G] we have Kerδ = N/[G,G] and Imδ = G/(N [G,G]). From γ : n[G,N ] 7→ n[G,G]we have Kerγ = N ∩ [G,G]/[G,N ] and Imγ = N/[G,G] = Kerδ. As deduced above, N/[G,N ] =(S/R)/[F/R, S/R] ∼= S/(R[F, S]) and so from β : s[F, S] 7→ s[F, S]R = sR[F, S] we have Kerβ =R ∩ [F, F ]/[F, S] and Imβ = S ∩ [F, F ]/(R[F, S]) ∼= N ∩ [G,G]/[G,N ] = Kerγ. Finally, from α :r[F,R] 7→ r[F, S] we have Imβ = R ∩ [F, F ]/[F, S] = Kerβ.

5.6(b): Applying part(a) to the group extension 1 → R → F G → 1 we obtain the exact se-quence

H2Fα //

∼=

H2Gβ// (H1R)G

γ//

∼=

H1Fδ //

∼=

H1G //

∼=

0

0 R/[F,R] F/[F, F ] G/[G,G]

where the first vertical isomorphism follows from Example II.4.1[1] and the second vertical isomorphismarises in the solution to part(a). By the First Isomorphism Theorem and exactness of the sequence,H2G ∼= H2G/Kerβ ∼= Imβ ∼= Kerγ = R ∩ [F, F ]/[F,R].

5.7(a): S3 is a closed orientable 3-manifold, and it has a group structure under quaternion multpli-cation (S3 < H as the elements of norm 1). The finite subgroup G of S3 provides a multiplication action,and it is free because the only solution in H to the equation gx = x for nontrivial x is g = 1. Froma result in the solution to Exercise I.4.2, the G-action is a “properly discontinuous” action and so the

16

quotient map ρ : S3 → S3/G is a regular covering space.Following a proof by William Thurston, S3/G is Hausdorff: considering two points α, β of S3 in distinctorbits, form respective neighborhoods Uα and Uβ such that they are disjoint and neither neighborhoodcontains any translates of α or β (this can be done by the Hausdorff property of S3). Taking the unionK of these two neighborhoods, Int(K −

⋃g 6=1 gK) yields a neighborhood of α and a neighborhood of β

which project to disjoint neighborhoods of Gα and Gβ in S3/G [note: we needed to refine K because ifUα intersects gUβ , then after projecting to the orbit space, ρ(gUβ) = ρ(Uβ) intersects ρ(Uα)].S3/G is a closed connected 3-manifold since it has S3 as a covering space (Theorem 26.5[6] provides thecompactness, Theorem 23.5[6] provides the connectedness, and the property of evenly-covered neighbor-hoods provides the nonboundary and manifold structure).Since the actions g : S3 → S3 are fixed-point free, we have deg(g) = (−1)3+1 = 1 by Theorem 21.4[4]and hence G acts by orientation-preserving homeomorphisms [note: orientation is in terms of local ori-entations µx ∈ H3(S3, S3−x) ∼= H3(S3) ∼= Z, as defined in [3] on pg234]. Local orientations µGx = µρ(x)

of S3/G = ρ(S3) are given by the images Γx(µx) = Γgx(µgx) under the isomorphisms of local homologygroups Γx : H3(S3, S3 − x) → H3(ρ(S3), ρ(S3) − ρ(x)) which arise from the Excision Theorem and thelocal-homeomorphism property of covering spaces; the ‘local consistency condition’ follows in the samerespect (where a ball B containing Gx and Gy has as preimage under ρ a union of balls, each of which ishomeomorphic to B, and such a homeomorphic ball containing g1x and g2y provides the local consistencycondition for S3). Therefore, S3/G is orientable.Since S3 is simply-connected, π1S

3 = 0 and so G ∼= π1(S3/G)/π1(S3) ∼= π1(S3/G). Applying PoincareDuality and the Universal Coefficient Theorem we obtainH2(S3/G) ∼= H1(S3/G) ∼= Hom(H1(S3/G),Z) ∼=Hom(Gab,Z) = 0 [noting that G is finite]. A theorem of Hopf (Theorem II.5.2[1]) gives us an exact se-quence which includes the surjection H2(S3/G)→ H2G→ 0, hence H2G = 0.

5.7(b): The binary icosahedral group G (of order 120) is the preimage in S3 which maps onto thealternating group A5 under S3 → SO(3) [up to isomorphism with the group of icosahedral-rotationalsymmetries] as explained in [3] on pg75. By part(a), H2G = 0. Consider the group extension 1→ K →G → A5 → 1 where K is the central kernel of order 2 (corresponding to G mapping onto A5). Theassociated 5-term exact sequence becomes 0 → H2(A5) → (H1K)A5

→ 0 because H1G ∼= Gab = 0,and thus we obtain the isomorphism H2(A5) ∼= (H1K)A5 . The A5-action on H1K ∼= Kab = K ∼= Z2

is induced by the conjugation G-action on K ∼= Z2 which is the trivial action (since K ≤ Z(G)), andtherefore H2(A5) ∼= H1K ∼= Z2.

5.7(c): Consider the abstract group G = 〈x, y, z ; x2 = y3 = z5 = xyz〉 which is a finite presenta-tion with the same number of generators as relations. We show that G is perfect (G = [G,G]) so thatH1G ∼= Gab = 0 and H2G = 0 by Exercise 5.5(b) above. Now G/[G,G] is an abelian group with therelations 2x = 3y = 5z = x + y + z. From this we see that x = y + z, so we need not look at x.Subsequently, 2y+ 2z = 3y ⇒ y = 2z and so we need not look at y. Finally, 5z = 3(2z) = 6z ⇒ z = 0and so all generators vanish, i.e. G/[G,G] = 0.Alternatively, the commutator subgroup is [G,G] ⊆ G, and to prove the opposite inclusion it suffices toshow that x, y, z all lie in [G,G] ≡ G′.

x2 = xyz ⇒︷︸︸︷x = yz ⇒ x2 = yzyz = y3 ⇒ z−1yzy−1 = z−2y ⇒

︷︸︸︷z2[z−1, y] = y

The two overbraced equations allow us to finish by showing that z ∈ G′.yzyz = y3 ⇒ y2 = zyz ⇒ [y, z] = z−1yz−1 ⇒ y = z[y, z]z

We then have z5 = xyz ⇒ z3 = xyz−1 = yz · z[y, z]z · z−1 = z[y, z]z · z2[y, z]⇒ z2 = [y, z]z3[y, z] =

z3z−3 · [y, z]z3[y, z]⇒ z−1 = g[y, z]⇒︷︸︸︷z = [y, z]−1g−1 ∈ G′ with g = z−3[y, z]z3 ∈ G′ by normality of

the commutator subgroup.

With the cyclic subgroup C = 〈xyz〉 we have G/C = A5 and hence the group extension 1→ C → G→A5 → 1 [A5 = 〈x, y, z ; x2 = y3 = z5 = xyz = 1〉 of order 60]. The associated 5-term exact sequencenow yields the isomorphism H2(A5) ∼= H1C = C, noting the trivial A5-action (since C is generated by acentral element) and noting the cyclicity Cab = C. Thus, by part(b) we deduce that |C| = 2 and hence|G| = 2 · 60 = 120. In fact, G = SL2(F5) is the binary icosahedral group!

17

6.1: Given N / G, let F be a projective resolution of Z over ZG and consider the complex FN . Since aprojective ZG-module is also projective as a ZH-module for any subgroup H ⊆ G (by Exercise I.8.2), Fis a projective resolution of Z over ZN and so H∗(FN ) = H∗N . By Exercise II.2.3(a), FN is a complexof G/N -modules and so H∗(FN ) inherits a G/N -action, with FN → FN given by x 7→ (gN)x = gx. ForCorollary II.6.3 we have the conjugation action α : N → N given by n 7→ gng−1 (for g ∈ G), and we havethe augmentation-preserving N -chain map τ : F → F given by x 7→ gx [it commutes with the boundaryoperator ∂ of F since ∂ is equivariant, and it satisfies the condition τ(nx) = gnx = gng−1gx = α(n)τ(x)].By Proposition II.6.2[1], if α is conjugation by g ∈ N then H∗(α) is the identity (hence trivial action),and so τ ′ : FN → FN is given by τ ′(x) = (gN)x = gx which agrees with the above map.

6.2: For any finite set A let Σ(A) be the group of permutations of A. For |A| ≤ |B|, choose an injectioni : A → B and consider the injection Σ(A) → Σ(B) obtained by extending a permutation on A to be theidentity on B − iA. In order to show that the induced map H∗Σ(A) → H∗Σ(B) is independent of thechoice of i, it suffices to show that any two injections i1 and i2 give conjugate maps i1, i2 : Σ(A) → Σ(B),because the conjugation map Σ(B)→ Σ(B) induces the identity map on homology H∗Σ(B)→ H∗Σ(B)by Proposition II.6.2[1]. Let τ be the permutation which takes i1(a) to i2(a) for all a ∈ A and is anarbitrary permutation (B − i1A) → (B − i2A). Then for a permutation i1(σ) = σ ∈ Σ(B), the per-mutation τστ−1 is equal to i2(σ). Thus i1 and i2 are conjugates of each other by τ , and the result follows.

6.3(a): Given the homomorphism α : G→ G′, the n-tuples in Cn(G) are sent to the n-tuples in Cn(G′)coordinate-wise via α, where [gh] 7→ [α(gh)] = [α(g)α(h)]. Thus H1(α) maps g to α(g), where g de-notes the homology class of the cycle [g]. We also have the explicit isomorphism H1G→ Gab given by g 7→gmod[G,G]. It is immediate that we have the commutative diagram H1G

H1(α)

∼= // Gab

α∗

H1G′ ∼= // G′ab

with α∗ : gmod[G,G] 7→ α(g) mod[G′, G′], which is precisely the map obtained from α by passage to thequotient. Thus, the isomorphism H1( ) ∼= ( )ab is natural.

6.3(b): Suppose G = F/R and G′ = F ′/R′ with F = F (S) and F ′ = F ′(S′) free, and supposeα : G → G′ lifts to α : F → F ′. Let Y and Y be associated to the presentation of G as in ExerciseII.5.3(a), and similarly for Y ′ and Y ′ with G′. Now α yields a map Y → Y ′ that sends the combinatorialpath s = S1

s to the combinatorial path α(s). ***Incomplete***

7.1: Consider G = G1 ∗A G2 with αk : A → Gk not necessarily injective, and let G1 = β1(G1),G2 = β2(G2), A = β1α1(A) = β2α2(A) be the images of G1, G2, A in G [where βk : Gk → G arises fromthe amalgamation diagram for G]. Form the amalgam H = G1 ∗A G2 and the commutative diagram:

A

i1

i2 // G2

j2

γ2

G1

γ1,,

j1// H

ϕ

Gwhere γk is the natural inclusion and ik is the obvious injection [subsequently, we have γ1i1 = γ2i2 andhence the unique map ϕ from the universal mapping property].

18

We also have the commutative diagram: A

α1

α2 // G2

β2

r2

G1

r1

β1

// G

φ

G2

j2

G1j1// H

where rk is βk with the codomain restricted to form the inclusion [subsequently, j1r1α1(a) = j1β1α1(a) =j1i1β1α1(a) = j2i2β1α1(a) = j2i2β2α2(a) = j2r2α2(a) and hence we have the unique map φ from theuniversal mapping property].From the diagrams (and dropping subscripts), φβ = jr and ϕj = γ.ϕφ(β(g)) = γ(r(g)) = γ(β(g)) = β(g)φϕ(j[β(g)]) = φγ[β(g)] = φ(β(g)) = j(r(g)) = j(β(g))Thus ϕφ = idG and φϕ = idH , and so ϕ = φ−1 is an isomorphism (H ∼= G).Consequently, any amalgamated free product is isomorphic to one in which the maps A→ Gk are injec-tive.

7.2:

7.3: Consider the Special Linear Group SL2(Z) ∼= Z4 ∗Z2Z6 which is the subgroup of all 2x2 inte-

gral matrices with determinant 1. Applying the Mayer-Vietoris sequence for groups and using the factthat H∗(Zk) is trivial in positive even dimensions and is isomorphic to Zk in positive odd dimensions,

we get the exact sequence 0→ H2n(SL2(Z))α→ Z2

β→ Z4 ⊕ Z6γ→ H2n−1(SL2(Z))→ 0.

Noting that the only nontrivial map ϕ : Z2 = 〈t〉 → Z4 = 〈s〉 is the canonical embedding defined byt 7→ s2, we assert that the induced map under H2n−1 is the same embedding: Considering the twoperiodic free resolutions of Z, there exists an augmentation-preserving chain map f between them byTheorem I.7.5[1] and we have a commutative diagram

Z[Z2]

fn+1

1+t// Z[Z2]

fn

t−1// Z[Z2]

Z[Z4]1+s+s2+s3

// Z[Z4]s−1// Z[Z4]

The left-side square yields (1+s+s2 +s3)fn+1(1) = fn(1+t) = fn(1)+ϕ(t)fn(1) = (1+s2)fn(1), and byexactness of the bottom row we have 0 = (s− 1)(1 + s2)fn(1) = (s3− s2 + s− 1)fn(1) ⇒ fn(1) = 1 + s,hence (1 + s+ s2 + s3)fn+1(1) = (1 + s2)(1 + s) = 1 + s+ s2 + s3 ⇒ fn+1(1) = 1. Then after movingto quotients, the cycle elements (for odd-dimensional homology) are mapped via ϕ∗(1) = 1 + 1 = 2while the boundary elements are mapped via ϕ∗(1) = 1, and the result follows [applies to all odd n, andf0(1) = 1].In general, an injection H → G = H × K onto a direct summand will pass to an injection under

any covariant functor T because the composition identity H → G H yields the identity T (H)j→

T (H ×K)→ T (H) which implies j is injective [this can be applied to H = Z2 and G = Z6 = Z2 × Z3].Thus we have β(t) = (s2

1, s32) injective and so H2n(SL2(Z)) ∼= Imα = Kerβ = 0. From the MV-sequence

we see that H2n−1(SL2(Z)) is a finite abelian group of order dividing |Z4 ⊕ Z6| = 24 = 23 · 3, hencecontains only 2-torsion and 3-torsion by the Primary Decomposition Theorem. Considering the 3-torsionin the MV-sequence, 0 → 0 → 0 ⊕ Z3 → H2n−1(SL2(Z))(3) → 0, we have H2n−1(SL2(Z))(3) = Z3 byexactness [this row can be extracted from the MV-sequence because finitely generated abelian groupsare direct sums of their Sylow pi-subgroups (by the Primary Decomposition Theorem) and maps be-tween the abelian groups will send primary components (the Sylow pi-subgroups) to respective primarycomponents]. For 2-torsion we consider the 2-torsion subgroup and its MV-sequence, and we obtain acommutative diagram

19

0

∼=

// Z2

∼=

// Z4 ⊕ Z2

∼=

// H2n−1(Z4 ∗Z2 Z2 = Z4) ∼= Z4

ψ

// 0

∼=

// 0

∼=

0 // Z2// Z4 ⊕ Z2

// H2n−1(SL2(Z))(2)// 0 // 0

Then by the Five-Lemma, ψ is an isomorphism and so H2n−1(SL2(Z))(2) = Z4.Thus, H2n−1(SL2(Z)) ∼= Z4 ⊕ Z3

∼= Z12.

⇒ Hi(SL2(Z)) ∼=

Z i = 0

Z12 i odd

0 i even

20

3 Chapter III: Homology and Cohomologywith Coefficients

0.1: Let F be a flat ZG-module and M a G-module which is Z-torsion-free (ie. Z-flat), and consider thetensor product F ⊗M with diagonal G-action. Since (F ⊗M)⊗G− = (F ⊗M ⊗−)G = F ⊗G (M ⊗−),it suffices to show that X = F ⊗G (M ⊗−) is an exact functor so that F ⊗M is ZG-flat (by Corollary10.5.41[2]). But by the same corollary M ⊗− is Z-exact (and a G-module) and F ⊗G − is ZG-exact, soX is exact and the result follows.

0.2: Let F be a projective ZG-module and M a Z-free G-module, and consider the tensor productF ⊗M with diagonal G-action. Since HomG(F ⊗M,−) ∼= Hom(F ⊗M,−)G ∼= Hom(F,Hom(M,−))G ∼=HomG(F,Hom(M,−)) where the second isomorphism is adjoint associativity (Theorem 10.5.43[2]), itsuffices to show that X = HomG(F,Hom(M,−)) is an exact functor so that F ⊗M is ZG-projective(by Corollary 10.5.32[2]). But by the same corollary Hom(M,−) is Z-exact (and a G-module) andHomG(F,−) is ZG-exact, so X is exact and the result follows.

1.1(a): For a finite group G and a G-module M we have the norm map N : MG → MG induced fromthe map M → M which is multiplication by the norm element N =

∑g∈G g. Noting that Nm = |G|m

for both m ∈ MG and m ∈ MG, we see that |G| · KerN = 0 (as Nm = Nm/∼ = 0 by definition ofkernel) and |G| · CokerN = 0 (as CokerN = MG/NM and NmmodNM = 0).

1.1(b): Suppose M is an induced module (M = ZG ⊗ A) where A is an abelian group and G actsby g · (r ⊗ a) = gr ⊗ a. Then MG = (ZG)G ⊗ A = Z ⊗ A and MG = (ZG)G ⊗ A = Z · N ⊗ A, whereN is the norm element. The norm map N : MG → MG is now given by z ⊗ a 7→ zN ⊗ a (for z ∈ Z)which is clearly a bijection. It is an isomorphism because N [z1⊗ a1 + z2⊗ a2] = Nz1⊗ a1 +Nz2⊗ a2 =z1N ⊗ a1 + z2N ⊗ a2 = N [z1 ⊗ a1] +N [z2 ⊗ a2].

1.1(c): Let M be a projective ZG-module; it is a direct summand of a free module F =⊕

i ZG. Byapplication of part(b) above with A =

⊕i Z we see that N is an isomorphism for F because F =

⊕i ZG =⊕

i(ZG⊗Z) = ZG⊗ (⊕

i Z). Now (M ⊕N)G ∼= Z⊗ZG (M ⊕N) ∼= (Z⊗ZGM)⊕ (Z⊗ZGN) ∼= MG⊕NG,and (M ⊕N)G = MG ⊕NG under the coordinate-wise G-action (of F) since g · (m,n) = (g ·m, g · n) =(m,n) ⇒ g ·m = m, g · n = n. Thus MG ⊕NG ∼= MG ⊕NG, and since the norm map is bilinear wehave MG

∼= MG.

1.2: Using the standard cochain complex, an element of C1(G,M) is a function f : G → M , andunder the coboundary map it is sent to (δf)(g, h) = g · f(h) − f(gh) + f(g). The kernel of this mapconsists of functions which satisfy f(gh) = f(g) + g · f(h), and these are derivations, so Z1(G,M) ∼=Der(G,M). Since an element of C0(G,M) is simply m ∈ M , and under the coboundary map it is sentto (δm)(g) = g · m − m, which is a principal derivation, we have B1(G,M) ∼= PDer(G,M). Thus,H1(G,M) ∼= Der(G,M)/PDer(G,M).IfG acts trivially onM then PDer(G,M) = 0 and Der(G,M) = Hom(G,M), soH1(G,M) ∼= Hom(G,M) =Hom(Gab,M) = Hom(H1G,M), where the second-to-last equality comes from the fact that any grouphomomorphism from G to an abelian group factors through the commutator subgroup [G,G] by Propo-sition 5.4.7[2]. In particular, H1(G) = 0 for any finite group G.

1.3: Let A be an abelian group with trivial G-action, and let F → Z be a projective resolutionof Z over ZG. Then F ⊗G A = (F ⊗ A)G ∼= Z ⊗G (F ⊗ A), and since the diagonal G-action onF ⊗ A is simply the left G-action on F (since G acts trivially on A), we can apply tensor associa-tivity (Theorem 10.4.14[2]) to obtain Z ⊗G (F ⊗ A) ∼= (Z ⊗G F ) ⊗ A ∼= FG ⊗ A. Thus there is auniversal coefficient sequence 0 → Hn(G) ⊗ A → Hn(G,A) → TorZ1 (Hn−1(G), A) → 0 by PropositionI.0.8[1]. Also, HomG(F,A) = Hom(F,A)G ∼= Hom(FG, A), where the last isomorphism arises because(gu)(m) = g · u(g−1m) = u(g−1m) and so we must have g−1m = m ∈ F for gu = u. Thus there isalso a universal coefficient sequence 0 → Ext1

Z(Hn−1(G), A) → Hn(G,A) → Hom(Hn(G), A) → 0 byProposition I.0.8[1].

21

1.4(a): “Let f : C ′ → C be a weak equivalence between arbitrary complexes, and let Q be a non-negative cochain complex of injectives. Then the map HomR(f,Q) : HomR(C,Q)→ HomR(C ′, Q) is aweak equivalence.”To prove this, note that the mapping cone C ′′ = C ⊕ ΣC ′ of f is acyclic by Proposition I.0.6[1], where(ΣC ′)p = C ′p−1 is the 1-fold suspension of C. The mapping cone of HomR(f,Q) is HomR(C ′′, Q) because

HomR(C ′, Q)n ⊕ ΣHomR(C,Q)n =∏q HomR(C ′q, Qq+n)⊕

∏q HomR(Cq, Qq+n−1) =∏

q HomR(C ′q ⊕ Cq+1, Qq+n) = HomR(C ′ ⊕ Σ−1C,Q)n = HomR(C ′′, Q)n

noting that C ′ ⊕ Σ−1C = (C ′q ⊕ Cq+1) = (C ′q−1 ⊕ Cq) = ΣC ′ ⊕ C = C ′′. Thus it suffices to show thatHomR(C ′′, Q) is acyclic (by Proposition I.0.6[1]), i.e., that Hn(HomR(C ′′, Q)) ≡ [C ′′, Q]n = 0 ∀ n ∈ Z.By the uniqueness part of the result of Exercise I.7.4, [C ′′, Q]n ≡ [ΣnC ′′, Q] is indeed 0, since any mapon Q is zero in negative dimensions (so all extensions off of that zero map are homotopy equivalent).

1.4(b): Let ε : F → Z be a projective resolution and let η : M → Q be an injective resolution.By part(a) above and noting that ε is a weak equivalence (regarded as a chain map with M concentratedin dimension 0), we have a weak equivalenceHomR(F,Q)← HomR(Z, Q). Similarly, by Theorem I.8.5[1]we have a weak equivalence HomR(F,M)→ HomR(F,Q).In particular, H∗(G,M) = H∗(QG) because H∗(G,M) = H∗(HomG(F,M)) and H∗(HomG(F,M)) ∼=H∗(HomG(Z, Q)) = H∗(Hom(Z, Q)G), noting that HomZ(Z, Q) ∼= Q.

2.1: Given projective resolutions F → M and P → N of arbitrary G-modules M and N , there is

an isomorphism of graded modules Γ : F ⊗G P∼=→ P ⊗G F given by f ⊗ p 7→ (−1)degf ·degpp ⊗ f , where

we consider diagonal G-action on Fi ⊗ Pj ∼= Pj ⊗ Fi [generally, degx = n for x ∈ Cn]. If we show that Γ(a degree 0 map) is a chain map, then it is a homotopy equivalence (hence a weak equivalence) and soTorG∗ (M,N) = H∗(F ⊗G P ) ∼= H∗(P ⊗G F ) = TorG∗ (N,M). Denote by d and d′ the boundary operatorsof F and P , respectively, and denote by D and D′ the boundary operators of F ⊗G P and P ⊗G F ,respectively. Then

D′Γ(f ⊗ p) = D′[(−1)degf ·degpp⊗ f ] = (−1)degf ·degp(d′p⊗ f + (−1)degpp⊗ df)= (−1)degf ·degpd′p⊗ f + (−1)degp·(degf+1)p⊗ df

andΓD(f ⊗ p) = Γ[df ⊗ p+ (−1)degff ⊗ d′p]

= (−1)degp·(degf−1)p⊗ df + (−1)degf+degf ·(degp−1)d′p⊗ f= (−1)degp·(degf+1)p⊗ df + (−1)degf ·degpd′p⊗ f

Thus D′Γ = ΓD and so Γ is a chain map.(This simultaneously provides a solution to Exercise I.0.5)

3.1: Let P be a projective R-module, and let C be a short exact sequence of S-modules which canbe regarded as R-modules via restriction of scalars. Since P is projective, HomR(P,C) is a short exact

sequence, and so it suffices to show that the isomorphism of functors HomS(S ⊗R P,−)∼=→ HomR(P,−)

is natural [because then HomS(S ⊗R P,C) is a short exact sequence which implies that S ⊗R P is aprojective S-module]. Given a module homomorphism ψ : M → N , we must check commutativity of thediagram

HomS(S ⊗R P,M)

ϕ1

α // HomS(S ⊗R P,N)

ϕ2

HomR(P,M)β

// HomR(P,N)

where α and β are given by f 7→ ψ f , and ϕi (i = 1, 2) is given by f 7→ f i under the universalmapping property with i : P → S ⊗R P , i(p) = 1 ⊗ p. Now ϕ2[α(F )] = ϕ2[ψ F ] = (ψ F ) i andβ[ϕ1(F )] = β[F i] = ψ (F i) = (ψ F ) i. Therefore, ϕ2α = βϕ1 and the result follows:Extension of scalars takes projective R-modules to projective S-modules.

3.2: Let Q be an injective R-module, and let C be a short exact sequence of S-modules which can be re-garded as R-modules via restriction of scalars. Since Q is injective, HomR(C, Q) is a short exact sequence,

22

and so it suffices to show that the isomorphism of functors HomS(−,HomR(S,Q))∼=→ HomR(−, Q) is

natural [because then HomS(C,HomR(S,Q)) is a short exact sequence which implies that HomR(S,Q)is an injective S-module]. Given a module homomorphism ψ : M → N , we must check commutativityof the diagram

HomS(N,HomR(S,Q))

ϕ1

α // HomS(M,HomR(S,Q))

ϕ2

HomR(N,Q)β

// HomR(M,Q)

where α and β are given by f 7→ f ψ, and ϕi (i = 1, 2) is given by f 7→ π f under the universalmapping property with π : HomR(S,Q)→ Q , π(f) = f(1). Now ϕ2[α(F )] = ϕ2[F ψ] = π (F ψ)and β[ϕ1(F )] = β[π F ] = (π F ) ψ = π (F ψ). Therefore, ϕ2α = βϕ1 and the result follows:Co-extension of scalars takes injective R-modules to injective S-modules.

3.3: Given S which is flat as a right R-module, let Q be an injective S-module, let C be a short exactsequence of S-modules, and consider HomR(C, Q) where C and Q are regarded as R-modules via re-striction of scalars. Since S is R-flat, S ⊗R C is a short exact sequence, and since Q is S-injective,HomS(S ⊗R C, Q) is a short exact sequence. It suffices to show that the isomorphism of functors

HomS(S ⊗R −, Q)∼=→ HomR(−, Q) is natural [because then HomR(C, Q) is a short exact sequence which

implies that Q is an injective R-module]. Given a module homomorphism ψ : M → N , we must checkcommutativity of the diagram

HomS(S ⊗R N,Q)

ϕ1

α // HomS(S ⊗RM,Q)

ϕ2

HomR(N,Q)β

// HomR(N,Q)

where α is given by f 7→ f (S⊗R ψ), β is given by f 7→ f ψ, ϕ1 is given by f 7→ f iN for the naturalmap iN : N → S ⊗R N , and ϕ2 is given similarly by f 7→ f iM . Now ϕ2[α(F )] = ϕ2[F (S ⊗R ψ)] =(F (S⊗R ψ)) iM = F ((S⊗R ψ) iM ) and β[ϕ1(F )] = β[F iN ] = (F iN ) ψ = F (iN ψ). Also,iN [ψ(m)] = 1 ⊗ ψ(m) = (S ⊗R ψ)(1 ⊗m) = (S ⊗R ψ)[iM (m)]. Therefore, ϕ2α = βϕ1 and the resultfollows:Restriction of scalars takes injective S-modules to injective R-modules if S is a flat right R-module.

3.4: Given S which is projective as a left R-module, let P be a projective S-module, let C be a shortexact sequence of S-modules, and consider HomR(P,C) where C and P are regarded as R-modules viarestriction of scalars. Since S is R-projective, HomR(S,C) is a short exact sequence, and since P isS-projective, HomS(P,HomR(S,C)) is a short exact sequence. It suffices to show that the isomorphism

of functors HomS(P,HomR(S,−))∼=→ HomR(P,−) is natural [because then HomR(P,C) is a short exact

sequence which implies that P is a projective R-module]. Given a module homomorphism ψ : M → N ,we must check commutativity of the diagram

HomS(P,HomR(S,M))

ϕ1

α // HomS(P,HomR(S,N))

ϕ2

HomR(P,M)β

// HomR(P,N)

where α is given by f 7→ φ f for φ(g) = ψ g (g : S → M), β is given by f 7→ ψ f , ϕ1 is givenby f 7→ πM f under the universal mapping property with π : HomR(P,M) → M , πM (f) = f(1),and ϕ2 is given similarly by f 7→ πN f . Now ϕ2[α(F )] = ϕ2[φ F ] = πN (φ F ) = (πN φ) Fand β[ϕ1(F )] = β[πM F ] = ψ (πM F ) = (ψ πM ) F . Also, πN [φ(f)] = πN [ψ f ] = (ψ f)(1) =ψ[f(1)] = ψ[πM (f)]. Therefore, ϕ2α = βϕ1 and the result follows:Restriction of scalars takes projective S-modules to projective R-modules if S is a projective left R-module.

4.1: Let R = Z/nZ, and note that the ideals of R are the ideals Ix = (x) mod(n) for x|n ∈ Z by the 4th

Isomorphism Theorem. It suffices to show that every map ϕ : I → R extends to a map R→ R so that Ris self-injective by Baer’s Criterion (Proposition III.4.1[1]). Given Ix and ϕ(x mod(n)) = r mod(n), wecan write n = yx so that ϕ(yx mod(n)) = 0. But ϕ(yx mod(n)) = yϕ(x mod(n)) mod(n) = yr mod(n)

23

and thus yr = mn = myx ⇒ r = xm. Then, since ϕ(x mod(n)) = [x mod(n)][m mod(n)] we canextend ϕ to the domain R by setting ϕ(1 mod(n)) = m mod(n).

(a): Let A be an abelian group such that nA = 0, and let C ⊆ A be a cyclic subgroup of order|C| = n. We can regard A as an R-module because xn = x0 ∈ R and R acts on A by xi · a = ia.As C is a [self-injective] subgroup of A, we have an inclusion C → A of an injective R-module intoan R-module. By definition of “injective module” (pg.782-783 of [5], statement XX.4.I1), every exactsequence of modules 0→ Q→M →M ′ → 0 splits for injective Q, hence Q is a direct summand of M .Therefore, C is a direct summand of A.

(b): Given A as above (i.e. an arbitrary abelian group of finite exponent), we regard A as an R-module. Since n is minimal (to annihilate A) there exists an element of order n and hence a cyclicsubgroup C ∼= R of order n in A. If we can show that A = A′ ⊕ A′′ for A′ of smaller exponent thenby induction on n we have that A′ is a direct sum of cyclic groups; thus it remains to show that A′′

is a direct sum of modules (each isomorphic to R) and is a direct summand of A. An application ofZorn’s Lemma on the set of direct sums with summands in A isomorphic to R (noting that the set isnonempty because it contains C) provides the maximal element A′′; we can use this lemma because anupper bound for any chain would be the direct sum of those elements (the direct sums) in that chain. Aring is Noetherian iff every ideal is finitely generated (by Theorem 15.1.2[2]); thus Z is Noetherian (beinga Principal Ideal Domain). Now R = Zn is also Noetherian, because a quotient of a Noetherian ring byan ideal is Noetherian (by Proposition 15.1.1[2]). It is a fact that a ring is Noetherian iff an arbitrarydirect sum of injective modules (over that ring) is injective. Thus A′′ =

⊕iR is R-injective, so A′′ is a

direct summand of A, and A is a direct sum of cyclic groups.This result is known as Prufer’s Theorem for abelian groups.

5.1: For any H-module M consider the G-module IndGHM =⊕

g∈G/H gM where this equality fol-

lows from Proposition III.5.1[1]. The summand gM is a gHg−1-module and hence gHg−1 is the isotropygroup of this summand in IndGHM . By Proposition III.5.3[1], IndGHM

∼= IndGgHg−1gM .In particular, by Proposition III.5.6[1] we have the K-isomorphism⊕

g∈E IndKK∩gHg−1ResgHg−1

K∩gHg−1gM ∼=⊕

gg′∈E IndKK∩gg′H(gg′)−1Resgg′H(gg′)−1

K∩gg′H(gg′)−1gg′M

Thus the K-module IndKK∩gHg−1ResgHg−1

K∩gHg−1gM depends up to isomorphism only on the class of g ∈ Ein K\G/H.

5.2(a): For any H-module M and G-module N consider the tensor product N ⊗ IndGHM which hasthe diagonal G-action. By Proposition III.5.1[1] and the fact that tensor products commute with directsums, we have N ⊗ IndGHM

∼= N ⊗ (⊕

g∈G/H gM) ∼=⊕

g∈G/H(N ⊗ gM) which has N ⊗M as a direct

summand in the underlying abelian group. Treating this as an H-module ResGHN ⊗M with a diago-nal action so that H is its isotropy group, we have N ⊗ IndGHM

∼= IndGH(ResGHN ⊗M) by PropositionIII.5.3[1].In particular, for M = Z we have N ⊗ Z[G/H] ∼= IndGHResGHN .

5.2(b): For any H-module M and G-module N consider U = Hom(IndGHM,N) which has the “di-agonal” G-action given by (gu)(m) = g · u(g−1m). By Proposition III.5.1[1] and the fact that theHom-functor “commutes” with direct sums/products, we have U ∼= Hom(

⊕gM,N) ∼=

∏Hom(gM,N),

where the indices on the sum/product symbols are implicitly the coset representatives in G/H. ThusU admits a direct product decomposition (πg : U Hom(gM,N)). Using the denotation πg(u) = ug,we have [(πgg0)(u)](m) = [πg(g0u)](m) = g0 · ug(g−1

0 m) = g0 · ug−10 g(m) = g0 · [πg−1

0 g(u)](m), and so

πgg0 ∼ πg−10 g, with m ∈ IndGHM . This decomposition has U Hom(M,N) as one of the surjections

in the underlying abelian group. Treating this surjection as an H-module π1 : U Hom(M,ResGHN)so that H is its isotropy group, we have Hom(IndGHM,N) ∼= CoindGHHom(M,ResGHN) by PropositionIII.5.8[1].An analogous proof will provide Hom(N,CoindGHM) ∼= CoindGHHom(ResGHN,M).

5.3: Let F be a projective G-module and M a Z-free G-module, and consider the tensor product F ⊗Mwith diagonal G-action. By Corollary III.5.7[1], ZG⊗M is a free G-module since M is free as a Z-module.

24

Since F is projective, it is a direct summand of a free G-module F =⊕

i ZG, so that F = F ⊕K. Asthe tensor product commutes with arbitrary direct sums (Corollary XVI.2.2[5]), F⊗M =

⊕i(ZG⊗M)

and hence is a free G-module. Finally, F⊗M = (F ⊕K)⊗M = (F ⊗M)⊕ (K ⊗M) and so F ⊗M isa direct summand of F⊗M , hence G-projective.Note that this gives a new proof of the result of Exercise III.0.2 above.

5.4(a): If |G : H| = ∞ then there are infinitely many distinct coset representatives in G/H. NowIndGHM =

⊕g∈G/H gM by Proposition III.5.1[1], and it has a transitive G-action which permutes the

summands. Consider an arbitrary [nontrivial] element x =∑Ni=1 gimi with all mi 6= 0 (this also refers

to the sum over all coset representatives with cofinitely many mi = 0). We may take the summandentry g1m1 of x and a summand g′M which doesn’t appear in the representation of x (i.e. g′ 6= gj for1 ≤ j ≤ N), and there then exists g′′ ∈ G such that g′′ · g1m1 = g′m because the G-action is transitiveon the summands. Thus x is not fixed by G (since g′′ · x 6= x), and so (IndGHM)G = 0.

Note that if the index is finite, then this result does not hold. For instance, take 2Z ⊂ Z = 〈x〉 whichhas index |Z : 2Z| = |Z2| = 2. Then IndZ

2ZM = M ⊕ xM where x is the coset representative of x(2Z)which generates Z2. Since x2i ·(m1, xm2) = (x2im1, x(x2im2)), we must have (IndZ

2ZM)Z ⊆M2Z⊕xM2Z.Since x2j+1 · (m1, xm2) = x · (m1, xm2) = (m2, xm1), we must have m1 = m2 and hence (IndZ

2ZM)Z =(m, xm) | m ∈ M2Z. For M = Z4 with 2Z-action defined as x2i ·m = (−1)im, the largest submoduleon which 2Z acts trivially [so that −m = m] is Z2. Thus (IndZ

2ZZ4)Z 6= 0 is our desired example.

5.4(b):

5.4(c): Assume statement (i), so that there is a finitely generated subgroup G′ ⊆ G such that |G′ : G′ ∩gHg−1| =∞ for all g ∈ G (with H ⊆ G). Using the analogue of Proposition III.5.6[1] for coinduction and

passing toG′-coinvariants, we obtain (ResGG′CoindGHM)G′ ∼= (⊕

g∈E CoindG′

G′∩gHg−1ResgHg−1

G′∩gHg−1gM)G′ ∼=⊕g∈E(CoindG

′

G′∩gHg−1M ′)G′ where this latter isomorphism follows from commutativity of the tensorproduct with direct sums (and E is the set of representatives for the double cosets KgH). It is afact that if G is finitely generated and |G : H| = ∞ then (CoindGHM)G = 0 for any H-module M[part(b) above], so by statement (i) we have (ResGG′CoindGHM)G′ =

⊕0 = 0 for any H-module M .

Noting that restriction of scalars gives the action g′ · x = ϕ(g′)x where ϕ is a map G′ → G, we have(CoindGHM)G ⊆ (ResGG′CoindGHM)G′ . Thus (CoindGHM)G = 0 and (i) implies (ii).Assume statement (ii), so that (CoindGHM)G = 0 for all H-modules M . Then in particular (for M = Z)there is only one element of (CoindGHZ)G, and that element must be zero, so (ii) implies (iii).Assume statement (iii), so that the element of (CoindGHZ)G represented by the augmentation mapε ∈ CoindGHZ is zero. We first note that [in general] if n0 = 0 ∈ NG for some n0 ∈ N then it isalso zero in NG′ for some finitely generated subgroup G′ ⊆ G; this is because NG = N/〈gn− n〉 and son0 can be written as a finite Z-linear combination of elements of the form gn − n, which implies thatwe can take G′ to be the subgroup generated by those specific g’s. In particular, ε = 0 ∈ (CoindGHZ)G′

for some finitely generated subgroup G′ of G. Using the double coset formula (analogue of PropositionIII.5.6[1]) and treating Z and other modules appropriately over specific groups (to ignore restriction),

we must have εg = 0 ∈ (CoindG′

G′∩gHg−1Z)G′ for all g ∈ G, where εg denotes the component of theaugmentation map in the specific summand of coinduction. Now if |G′ : G′ ∩ gHg−1| < ∞ then

CoindG′

G′∩gHg−1Z ∼= IndG′

G′∩gHg−1Z ∼= Z[G′/G′ ∩ gHg−1], giving εg =∑g′∈K g

′ ⊗ εg(g′) = (

∑g′) ⊗ 1

where K is the set of coset representatives for the quotient G′/G′ ∩ gHg−1. Then g′′εg = εg ∀ g′′ ∈ G′,so εG 6= 0 ∈ (CoindG

′

G′∩gHg−1Z)G′ . Thus we must have |G′ : G′ ∩ gHg−1| =∞, so (iii) implies (i).

5.5: Let G be a finite group and let k be a field, and consider the free module kG. We have kG ∼=kG⊗k k = IndG1k

∼= CoindG1k = Homk(kG, k), where the second-to-last equation follows from the ana-logue over k of Proposition III.5.9[1] since |G| <∞. Then HomkG(−, kG) ∼= HomkG(−,Homk(kG, k)) ∼=Homk(−, k) where the last isomorphism follows from the universal property of co-induction. It is a factthat every k-vector space is an injective k-module [if the vector space W with basis B is a subspace of avector space V , then we can extend B to a basis of V and then V = W ⊕U where U is the vector spacespanned by the additional basis vectors extended from B]; thus k is injective ⇒ Homk(−, k) is exact ⇒HomkG(−, kG) is exact ⇒ kG is self-injective as a kG-module.

25

A Noetherian ring is a commutative ring which satisfies the Ascending Chain Condition on ideals (i.e.no infinite increasing chain of ideals), and any field k is Noetherian because the only ideals are 0 andk (giving 0 ⊆ k) by Proposition 7.4.9[2]. Now kG is Noetherian (for G finite) because it is a finite-dimensional k-vector space and so any infinite ascending chain of subspaces would require cofinitely manyof those subspaces to have dimension greater than |G|, a contradiction; since ideals of kG are necessarilyk-subspaces, the result follows. It is a fact that a ring R is Noetherian iff an arbitrary direct sum ofinjective R-modules is injective. Thus the free module F =

⊕i kG is injective and so any projective

kG-module is kG-injective (because a projective module is a direct summand of a free module, and adirect summand of an injective module is injective).Assuming the claim is true that any kG-module is a submodule of a kG-projective module, then bydefinition of “injective” it follows that any injective kG-module is a direct summand of a kG-projectivemodule, hence kG-projective; it remains to prove this claim. We have a canonical kG-module injectivemap M → HomkH(kG,M) where the kG-module M can be regarded as a kH-module by restriction ofscalars (see pg64 of [1]). But HomkH(kG,M) = CoindGHM

∼= IndGHM = kG⊗kH M , where this second-to-last equation follows from the analogue over k of Proposition III.5.9[1] since |G| <∞. Using H = 1,this says that M is a submodule of kG ⊗k M . But M is treated as a k-vector space (M ∼=

⊕i k), so

kG⊗kM ∼=⊕

i(kG⊗k k) ∼=⊕

i kG; i.e. kG⊗kM is a free [hence projective] kG-module.

6.1(a): We first note that an arbitrary direct sum of projective resolutions is projective, which followsfrom the fact that an arbitrary direct sum of projective modules is projective and from the exactness ofthe row for each summand. We then note that homology commutes with direct sums, and this followsfrom the obvious facts Ker(

⊕i di) =

⊕i Ker(di) and Im(

⊕i di) =

⊕i Im(di). From these two facts and

noting that the tensor product commutes with direct sums, we see that the Tor-functor commutes withdirect sums,

TorR∗ (M,⊕

iNi) = H∗(M ⊗R (⊕

iNi))∼= H∗(

⊕i(M ⊗R Ni)) ∼=

⊕iH∗(M ⊗R Ni) =

⊕i TorR∗ (M,Ni).

For the amalgamation G = G1 ∗A G2 consider the short exact sequence of permutation modules 0 →Z[G/A]→ Z[G/G1]⊕ Z[G/G2]→ Z→ 0. By Proposition III.6.1[1] we have the long exact sequence

// Hn(G,Z[G/A])

∼=

// Hn(G,Z[G/G1]⊕ Z[G/G2])

∼=

// Hn(G,Z)

∼=

//

Hn(A) Hn(G1)⊕Hn(G2) Hn(G)

where the vertical isomorphisms follow from the fact H∗(H) = H∗(G,Z[G/H]), and the middle isomor-phim utilizes commutativity of direct sums which follows from above because H∗(G,−) = TorG∗ (Z,−).This long exact sequence is the Mayer-Vietoris sequence for the amalgam G.

6.1(b): For the amalgamation G = G1 ∗AG2 consider the short exact sequence of permutation modules0→ Z[G/A]→ Z[G/G1]⊕Z[G/G2]→ Z→ 0. Applying −⊗M still yields a short exact sequence becauseall of the modules are free Z-modules (we consider the sequence of permutation modules as a free reso-lution of Z, hence a homotopy equivalence by Corollary I.7.6[1]); this homotopy equivalence for Z givesa homotopy equivalence for Z ⊗M = M (since functors preserve identities), hence a weak equivalence(which can be stated as a resolution). By Proposition III.5.6[1] we have Z[G/A] ⊗M ∼= IndGAResGAM ,and by Shapiro’s Lemma we apply H∗(G,−) to obtain H∗(G, IndGAResGAM) ∼= H∗(A,ResGAM). Similarresults follow for the other modules, and so by Proposition III.6.1[1] the exact sequence (which resultedfrom the sequence of permutation modules after application of −⊗M) yields a long exact sequence [theMayer-Vietoris sequence for homology with coefficients]

Hn(A,ResGAM) // Hn(G1,ResGG1M)⊕Hn(G2,ResGG2

M) // Hn(G,M)

Now consider the original sequence of permutation modules, but instead apply Hom(−,M) which stillyields a short exact sequence (same reason as mentioned above). By the analogue over co-induction ofProposition III.5.6[1] (or the result of Exercise III.5.2(a) above) we have Hom(Z[G/A],M) ∼= CoindGAResGAM ,and by Shapiro’s Lemma we apply H∗(G,−) to obtain H∗(G,CoindGAResGAM) ∼= H∗(A,ResGAM). Similarresults follow for the other modules, and so by Proposition III.6.1[1] the exact sequence (which resultedfrom the sequence of permutation modules after application of Hom(−,M) yields a long exact sequence[the Mayer-Vietoris sequence for cohomology with coefficients]

26

Hn(G,M) // Hn(G1,ResGG1M)⊕Hn(G2,ResGG2

M) // Hn(A,ResGAM)

7.1(a): Consider the exact sequence · · · → C1∂1→ C0

ε→ M → 0, where each Ci is H∗-acyclic. Wecan apply the dimension-shifting technique using the short exact sequences 0 → Kerε → C0 → M → 0and 0 → Ker∂i → Ci → Ker∂i−1 → 0 to obtain the isomorphism Hn(G,M) ∼= H1(G,Ker∂n−2) ∼=KerH0(G,Ker∂n−1) → H0(G,Cn−1) = Ker(Ker∂n−1)G → (Cn−1)G. Now consider the diagram be-low concerning CG

(Cn+1)G∂n+1

//

γ1

%% %%

(Cn)G∂n //

β

%% %%

(Cn−1)G

(Zn)G

α

::

(Zn−1)G

γ299

with Zi = Ker∂i, noting that the composition βα is exact [right-exactness of (−)G on 0 → Ker∂n →Cn → Ker∂n−1 → 0] and Kerγ2

∼= Hn(G,M). Thus Im∂n+1 = Imα = Kerβ (since γ1 is surjective) and soHn(CG) = Ker∂n/Kerβ. Noting that (Zn−1)G = (Cn)G/Kerβ by the First Isomorphism Theorem, takethe kernel of γ2 [denoted K/Kerβ] and take its preimage under β to obtain K in (Cn)G. Since K mapsto zero under γ2β = ∂n we have K ⊆ Ker∂n. We also have Ker∂n ⊆ K [hence they are equal] becauseby commutativity of the maps Ker∂n maps to K/Kerβ and hence lies in K. Thus Ker∂n/Kerβ = Kerγ2

and so H∗(G,M) ∼= H∗(CG).

7.1(b): Consider the exact sequence 0 → M → C0 δ0→ C1 δ1→ · · · , where each Ci is H∗-acyclic. Wecan apply the dimension-shifting technique using the short exact sequences 0→M → C0 → Kerδ0 → 0and 0 → Kerδi−1 → Ci → Kerδi → 0 to obtain the isomorphism Hn(G,M) ∼= H1(G,Kerδn−2) ∼=CokerH0(G,Cn−1) → H0(G,Kerδn−1) = Coker(Cn−1)G → (Kerδn−1)G. Using a similar approachas in part(a) above, we see that H∗(G,M) ∼= H∗(CG).

7.2: This will reprove Proposition III.2.2[1] on isomorphic functors.Method 1 : Let M be Z-torsion-free, so that M ⊗ − is an exact functor (M is Z-flat). Then

H∗(G,M ⊗ −) is a homological functor because given a short exact sequence of modules C, M ⊗ Cis a short exact sequence and F ⊗G (M ⊗ C) is a short exact sequence of chain complexes (F is projec-tive, hence flat), so the corresponding long exact homology sequence gives us the desired property (byLemma 24.1[4] and Theorem 24.2[4]). Similarly, TorG∗ (M,−) = H∗(F

′ ⊗G −) is a homological functor,where F ′ → M is a projective resolution. Both functors are effaceable [erasible] in positive dimensions,since the chain complexes F ⊗G (M ⊗ P ) and F ′ ⊗G P are exact for P projective. In dimension 0,H0(G,M ⊗ N) ∼= Z ⊗G (M ⊗ N) ∼= (M ⊗ N)G = M ⊗G N ∼= TorG0 (M,N). Therefore, by TheoremIII.7.3[1] we have an isomorphism of ∂-functors H∗(G,M ⊗−) ∼= TorG∗ (M,−). [The case for cohomologyis similar].

Method 2 : The chain complex associated to the group TorG∗ (M,N) is given by · · · → F ′0 ⊗G N →M⊗GN → 0, where F ′ →M is a projective resolution. This can be rewritten as · · · → (F ′0⊗N)G → (M⊗N)G → 0 which yields TorG∗ (M,N) = H∗(CG), where C is the chain complex (F ′i ⊗N). This is indeed anexact sequence because the universal coefficient theorem yields H∗(F

′⊗N) = H∗(F′)⊗N = N (nontrivial

only in dimension 0). Now F ′i is projective and hence a summand of a free module F = F ′i ⊕K =⊕

j ZG.Then H∗(G,F⊗N) ∼= H∗(G,F

′i⊗N)⊕H∗(G,K⊗N) and H∗(G,F⊗N) ∼=

⊕j H∗(G,ZG⊗N) = 0, noting

that induced modules are H∗-acyclic by Corollary III.6.6[1]. Thus H∗(G,F′i ⊗N) = 0 and each F ′i ⊗N is

H∗-acyclic. We can now apply Exercise III.7.1(a) which implies TorG∗ (M,N) = H∗(CG) ∼= H∗(G,M⊗N).[The case for cohomology is similar].

7.3: For dimension-shifting in homology, we can choose the induced module M = ZG ⊗Z M whichmaps onto M by ϕ(r⊗m) = rm; it is an H∗-acyclic module by Corollary III.6.6[1]. This map is Z-splitbecause it composes with the natural map i : M →M to give the identity, m 7→ 1⊗m 7→ 1m = m.

For dimension-shifting in cohomology, we can choose the coinduced module M = HomZ(ZG,M) which

provides the embedding M → M given by m 7→ (r 7→ rm); it is an H∗-acyclic module by Corollary

III.6.6[1]. This map is Z-split because it composes with the natural map π : M →M to give the identity,m 7→ (r 7→ rm) 7→ [r 7→ rm](1) = 1m = m.

27

8.1: Let H be a central subgroup of G and let M be an abelian group with trivial G-action. Thenthe isomorphism c(g) : (H,M)→ (gHg−1,M) becomes the identity on (H,M) given by (h 7→ ghg−1 =h,m 7→ gm = m). By Corollary III.8.2[1], this conjugation action of G on (H,M) induces an action ofG/H on H∗(H,M) given by gH · z = c(g)∗z. Letting α : H → gHg−1 = H denote the group map ofc(g), and letting F be a projective resolution of Z over ZG, we can choose the chain map τ : F → Fto be the identity, since it satisfies the condition τ(hx) = hx = α(h)x = α(h)τ(x). Thus the chain mapF ⊗HM → F ⊗HM given by x⊗m 7→ x⊗ gm = x⊗m is the identity, and so the induced map c(g)∗ isthe identity on H∗(H,M) which gives the trivial G/H-action gH ·z = c(g)∗z = z. Similarly, by CorollaryIII.8.4[1], we have an induced action of G/H on H∗(H,M) given by gH ·z = (c(g)∗)−1z. Using the samechain map τ , we have the cochain map HomH(F,M)→ HomH(F,M) given by f 7→ [x 7→ gf(x) = f(x)]which is the identity. Thus the induced map c(g)∗ is the identity on H∗(H,M) which gives the trivialG/H-action gH · z = (c(g)∗)−1z = z.Alternatively, the trivial G/H-action follows immediately from the fact that functors preserve identities,where H∗ and H∗ are the functors in question and c(g) is the identity map in question.

8.2: Let α : H → G be an inclusion, let M be an H-module, let i : M → IndGHM be the canon-ical H-map i(m) = 1 ⊗ m, and let π : CoindGHM → M be the canonical H-map π(f) = f(1). Wecan take the chain map τ : F → F to be the identity (F is a free resolution of Z over ZG) sinceτ(hx) = hx = α(h)x = α(h)τ(x) and F can be regarded as a free resolution over ZH.Consider (α, i)∗ on the chain level, induced by the map F ⊗H M → F ⊗G IndGHM given by x ⊗m 7→x ⊗ (1 ⊗ m). This map is a homotopy equivalence because we can use the universal property of ten-sor products to define its inverse x ⊗ (g ⊗ m) = xg ⊗ (1 ⊗ m) 7→ xg ⊗ m, the composite map beingx ⊗ (g ⊗ m) 7→ xg ⊗ m 7→ xg ⊗ (1 ⊗ m) = x ⊗ g · (1 ⊗ m) = x ⊗ (g ⊗ m). In particular we have aweak equivalence which yields the isomorphism H∗(H,M) ∼= H∗(G, IndGHM) of Shapiro’s Lemma givenby (α, i)∗.Now consider (α, π)∗ on the cochain level, induced by the map HomH(F,M) → HomG(F,CoindGHM)given by [x 7→ f(x)] 7→ [x 7→ (g 7→ gf(x))] with g ∈ ZG. This map is a homotopy equivalence becausewe can use the universal property of co-induction to define its inverse [x 7→ (g 7→ gf(x))] 7→ [x 7→(1 7→ f(x))] = [x 7→ f(x)]. In particular we have a weak equivalence which yields the isomorphismH∗(H,M) ∼= H∗(G,CoindGHM) of Shapiro’s Lemma given by (α, π)∗.

9.1: Considering homology, let x ⊗H m represent z ∈ H∗(H,M). Computing corGHz on the chainlevel yields x⊗H m 7→ x⊗G m, while computing corGgHg−1gz on the chain level yields gx⊗gHg−1 gm 7→gx⊗G gm = x⊗G m. Since the images are equal, corGgHg−1gz = corGHz.

Considering homology, let x ⊗G m represent z ∈ H∗(G,M). Computing resGgHg−1z on the chain level

yields x ⊗G m 7→∑g′∈gHg−1\G g

′ · (x ⊗gHg−1 m). Computing g · resGHz on the chain level yields

x⊗Gm 7→∑g′∈H\G g

′ ·(x⊗Hm) 7→∑g′∈H\G(gg′)·(x⊗gHg−1m) =

∑g′′∈gHg−1\G g

′′ ·(x⊗gHg−1m), where

this last equality arises from g(Hg′) = gHg−1gg′ = (gHg−1)g′′ with g′′ = gg′ being the coset representa-tive. Since the images are equal (the sums are the same for all coset representatives), g·resGHz = resGgHg−1z.

Considering cohomology, let fG represent z ∈ H∗(G,M). Computing g · resGHz on the chain level yields[x 7→ fG(x)] = [x 7→ g−1fG(gx)] 7→ [x 7→ g−1fH(gx)] 7→ [x 7→ gg−1fgHg−1(g−1gx)] = [x 7→ fgHg−1(x)],while computing resGgHg−1z on the chain level yields [x 7→ fG(x)] 7→ [x 7→ fgHg−1(x)]. Since the images

are equal, g · resGHz = resGgHg−1z.

Considering cohomology, let f represent z ∈ H∗(H,M). Computing corGHz on the chain level yieldsf 7→

∑g′∈G/H g

′ · f , where G acts diagonally on f ∈ Hom(F,M). Computing corGgHg−1gz on the

chain level yields g · f 7→∑g′∈G/gHg−1(g′g) · f =

∑g′′∈G/H g

′′ · f , where this last equality arises from

g′(gHg−1)g = g′gH = g′′H with g′′ = g′g being the coset representative. Since the images are equal(the sums are the same for all coset representatives), corGgHg−1gz = corGHz.

9.2: The transfer map H1(G) → H1(H) can be regarded as a map of abelian groups Gab → Hab.If g denotes the representative of Hg then ρg = gg−1, where ρ : G→ H is the unique map of left H-setswhich sends ever coset representative to 1. Now the transfer map is induced by the composite chain map

F (G)Gtr→ F (G)H → F (H)H , where the latter map concerns the chain map τ : F (G) → F (H) given

by (g0, g1) 7→ (ρg0, ρg1). Using bar notation, this composite chain map is given by [g] 7→∑g′∈E g

′[g] 7→

28

∑g′∈E τ(g′, g′g) =

∑g′∈E(g′g′

−1, ρ(g′g)) =

∑g′∈E(1, ρ(g′g)) =

∑g′∈E [ρ(g′g)], where E is a set of rep-

resentatives for the right cosets Hg′ and we note then that g′ = g′. Note that for homology classes,[g1]+[g2] = [g1g2] because of the boundary map ∂2[g1|g2] = [g2]− [g1g2]+[g1]; thus (as a homology class)∑g′∈E [ρ(g′g)] = [

∏g′∈E ρ(g′g)]. Using the isomorphism H1(G)→ Gab given by [g] 7→ g mod [G,G], the

transfer map Gab → Hab is computed as g mod [G,G] 7→∏g′∈E g

′g(g′g)−1 mod [H,H].

Example: The transfer map Z→ nZ is multiplication by n, sincex 7→ [1x(1x)−1][xx(xx)−1] · · · [xn−2x(xn−2x)−1][xn−1x(xn−1x)−1] = 1 · 1 · · · 1 · xn(1)−1 = xn, where x is

the generator of Z.

10.1: The symmetric group G = S3 on three letters is the group of order 3! = 6 whose elements are thepermutations of the set 1, 2, 3. The Sylow 3-subgroup is generated by the cycle (1 2 3), and a Sylow2-subgroup is generated by the cycle (1 2). Noting the semi-direct product representation S3 = Z3 o Z2

where Z2 acts on Z3 by conjugation, we have H∗(S3) = H∗(S3)(2) ⊕H∗(S3)(3)∼= H∗(S3)(2) ⊕H∗(Z3)Z2

by Theorem III.10.3[1]. Now S3 is the unique nonabelian group of order 6, so D6∼= S3 and we can use

Exercise AE.9 which implies that the Z2-action on H2i−1(Z3) ∼= H2i(Z3) is multiplication by (−1)i (wecan pass this action to cohomology by naturality of the UCT). Thus Hn(Z3)Z2 is isomorphic to Z3 forn = 2i where i is even, and is trivial for n odd and n = 2i where i is odd. Taking any Sylow 2-subgroupH ∼= Z2, Theorem III.10.3[1] states that H∗(S3)(2) is isomorphic to the set of S3-invariant elements ofH∗(H). In particular we have the monomorphism H2i−1(S3)(2) → H2i−1(H) = 0, so H2i−1(S3)(2) = 0.

An S3-invariant element z ∈ H2i(H) ∼= Z2 must satisfy the equation resHKz = resgHg−1

K gz, where Kdenotes H ∩ gHg−1. If g ∈ H then gHg−1 = H and the above condition is trivially satisfied for all z(hz = z by Proposition III.8.1[1]). If g /∈ H then K = 1 because H is not normal in S3 and onlycontains two elements, so the intersection must only contain the trivial element. But then the image ofboth restriction maps is zero, so the condition is satisfied for all z; thus H2i(S3)(2) = Z2. Alternatively,a theorem of Richard Swan states that if G is a finite group such that Sylp(G) is abelian and M is a

trivial G-module, then Im(resGSylp(G)) = H∗(Sylp(G),M)NG(Sylp(G)). It is a fact that NS3(Z2) = Z2 (refer

to pg51[2]), so taking G = S3 and H = Syl2(S3) ∼= Z2 and M = Z we have Im(resS3

H ) = (Z2)Z2 = Z2 inthe even-dimensional case. Since any invariant is in the image of the above restriction map (by TheoremIII.10.3[1]), the result H2i(S3)(2) = Z2 follows.

⇒ Hn(S3) ∼=

Z n = 0

Z6 n ≡ 0 mod 4 , n 6= 0

Z2 n ≡ 2 mod 4

0 otherwise

10.2(a): Let H be a subgroup of G of finite index, let C be the double coset HgH, and let T (C) be the en-domorphism H∗(G, f(C)) of H∗(H,M) where f(C) is the G-endomorphism of IndGHM given by 1⊗m 7→∑c∈C/H c

−1 ⊗ cm. To show that T (C)z = corHH∩gHg−1resgHg−1

H∩gHg−1gz, it suffices to check this equation

in dimension 0 (by Theorem III.7.5[1]). The right side maps m ∈ MH to∑h∈H/H∩gHg−1 h(gm) =∑

hg∈HgH/H(hg)m =∑g′∈C/H g

′m ∈ MH , where g′ = hg as a coset representative. Now T (C) indimension 0 is given by the composite map

H0(H,M)α→ H0(G,CoindGHM)

β→ H0(G, IndGHM)f∗→ H0(G, IndGHM)

β−1

→ H0(G,CoindGHM)α−1

→H0(H,M)

where α is the Shapiro isomorphism m 7→ (s 7→ m), and β is the canonical isomorphism F 7→∑x∈G/H x⊗

F (x−1), and f∗ is induced by f(C), and α−1 is the inverse F 7→ F (1), and β−1 is the inverse x⊗m 7→β−1[x⊗m](s · x) which is sxm if sx ∈ H and is 0 if sx /∈ H. This composite is given by

T (C) : m 7→ (s 7→ m) 7→∑x∈G/H x⊗m 7→

∑x

∑c∈C/H x(c−1 ⊗ cm) =

∑x

∑c xc−1 ⊗ cm 7→∑

x

∑c β−1[xc−1 ⊗ cm](s · xc−1) 7→

∑x

∑c β−1[xc−1 ⊗ cm](xc−1) ∈MH

To simplify this last term, note that the image of β−1[xc−1 ⊗ cm] is nontrivial iff xc−1 ∈ H ⇔ x ∈C, and for each x there is at most one c such that xc−1 ∈ H. Thus the double sum reduces to

29

∑x∈G/H , x∈C xc

−1 · cm =∑x∈C/H xm, and this is precisely the image of the right-side map.

Note: α was determined by noting that any element f of (CoindGHM)G must satisfy f(xg) = f(x) andhence f is determined by f(1) = m. So f is given by g 7→ m, but it must also commute with the H-actionwhich means that hg 7→ hm and hence hm = m, i.e. m ∈MH . Thus α(m) = (s 7→ m), s ∈ G.

10.2(b): If z ∈ H∗(H,M) is G-invariant where H ⊆ G is of finite index as above, then T (C)z =

corHH∩gHg−1resgHg−1

H∩gHg−1gz = corHH∩gHg−1resHH∩gHg−1z = |H : H ∩ gHg−1| = a(C)z, where the second-to-

last equality follows from Proposition III.9.5[1].

10.2(c): Let X = z ∈ H∗(H,M) | T (C)z = a(C)z ∀ C where C is any H-H double cosetand a(C) = |C/H| = |H : H ∩ gHg−1|. Since the image of the restriction map resGH lies in theset of G-invariant elements of H∗(H,M), and such elements lie in X by part(b) above, we haveIm(resGH) ⊆ X. In the situation of Theorem III.10.3[1] and Proposition III.10.4[1], consider the ele-ment w = corGHz ∈ Hn(G,M) where z is an arbitrary element of X. Then either Hn(H,M) is anni-hilated by |H| [H = Sylp(G)] in which case w ∈ Hn(G,M)(p), or |G : H| is invertible in M [hence

in Hn(G,M)]. Using Proposition III.9.5[1] we obtain resGHw =∑g∈H\G/H corHH∩gHg−1resgHg

−1

H∩gHg−1gz =∑g∈H\G/H T (C)z =

∑g∈H\G/H a(C)z = (

∑g∈H\G/H |H : H ∩ gHg−1|)z = |G : H|z. Since either

|G : H| is prime to p or is invertible in M , it follows that z = resGHw′ where w′ = w/|G : H|. Thus

X ⊆ Im(resGH) ⇒ X = Im(resGH).

30

4 Chapter IV: Low-Dimensional Cohomologyand Group Extensions

2.1: If d is a derivation [crossed homomorphism], then d(1) = d(1 · 1) = d(1) + 1 · d(1) = 2 · d(1) and sod(1) = 0.

2.2: Let I be the augmentation ideal of ZG and let D : G→ I be the derivation defined by g 7→ g−1 (thisis the principal derivation G→ ZG corresponding to 1 ∈ ZG). Given any G-module A and any derivationd : G→ A, we can extend d to an additive map d : ZG→ A such that d(rs) = dr ·ε(s) + r · ds, where ε isthe augmentation map and r, s ∈ ZG. This map is well-defined because d(rs · t) = d(rs) · ε(t) + rs · dt =dr · ε(s) · ε(t) + r · ds · ε(t) + rs · dt = dr · ε(st) + r · (ds · ε(t) + s · dt) = dr · ε(st) + r · d(st) = d(r · st),and dg = d(g1) = dg · ε(1) + g · d1 = dg · 1 + g · d1 = dg + g · 0 = dg. The restriction f of d to I isa G-module homomorphism since f(r · s) = f(r) · ε(s) + r · f(s) = f(r) · 0 + r · f(s) = r · f(s), andf(g − 1) = dg − d1 = dg − d1 = dg − 0 = dg, so f is the unique module map I → A such that d = fD.This means D is the universal derivation on G, and Der(G,A) ∼= HomZG(I, A).

2.3(a): Let F = F (S) be the free group generated by the set S, and consider the F -module A witha family of elements (as)s∈S . For the set map S → A o F defined by s 7→ (as, s), there is a uniqueextension to a homomorphism ϕ : F → AoF by the universal mapping property of F . This is a splittingof 1 → A → A o F

π→ F → 1 because πϕ(f) = π(df, f) = f , where d : F → A is some function whichmaps s ∈ S to as ∈ A. Since derivations F → A correspond to splittings of the above group extension,d is the unique derivation such that ds = as ∀ s ∈ S.

2.3(b): Given any function φ : S → A where A is a ZF -module, there is a unique map d : F → A bypart(b) above, hence a unique ZF -module homomorphism f : I → A such that fD = d by ExerciseIV.2.2 above. In particular, I satisfies the universal property of free modules, φ = fD|S , and so theaugmentation ideal I of ZF is a free ZF -module with basis (Ds)s∈S = (s− 1)s∈S .Note that this reproves Exercise II.5.3(b).

2.3(c): By part(b) above the universal derivation D : F → I satisfies Df =∑s∈S(∂f/∂s)ds, where

I is the augmentation ideal of ZF . By Exercise IV.2.2 any derivation d : F → M (where M is anF -module) corresponds to a unique F -module map ϕ : I → M and hence satisfies df = ϕ(Df) =∑s∈S(∂f/∂s)ϕ(Ds) =

∑s∈S(∂f/∂s)ds, where ∂f/∂s lies in ZF .

Note that this reproves Exercise II.5.3(c).

2.4(a): Let G = F/R where F = F (S) and R is the normal closure of some subset T ⊆ F . Forany G-module A, derivations d : G→ A correspond to splittings of 1→ A→ AoG→ G→ 1; they areof the form s(g) 7→ (dg, g) ∈ A oG. Consider the homomorphism ϕ : F → A oG given by f 7→ (df, g)where g is the image of f under the projection map p : F → F/R, which is the extension of the set mapS → A o G, s 7→ (ds, p(s)), by the universal mapping property of F . Now r = ftf−1 ∈ R is mappedto ϕ(r) = df + f · (dt + t · df−1) = df + f · dt − ftf−1 · df = (1 − r) · df + f · dt = dt, so ϕ induces ahomomorphism G→ AoG iff dt = 0 ∀ t ∈ T [note: 1− r = 1− 1 = 0 when computing the G-action ondf ∈ A]. This homomorphism is a splitting iff d is a derivation, and so derivations G→ A correspond toderivations d : F → A such that d(T ) = 0.

2.4(b): From part(a) above and Exercise IV.2.3(c) we see that derivations G → A correspond toderivations d : F → A such that dt =

∑s∈S(∂t/∂s)ds = 0 for all t ∈ T , where ∂t/∂s is the image of

∂t/∂s under ZF → ZG due to the G-action on A (restriction of scalars). By Exercise IV.2.3(a) thesecorrespond to families (as)s∈S of elements of A such that

∑s∈S(∂t/∂s)as = 0 for all t ∈ T .

2.4(c): The identity map idI : I → I [where I is the augmentation ideal of ZG] corresponds to aderivation d : G → I such that ds = idID(s) = idI(s − 1) = s − 1 by Exercise IV.2.2, and this cor-responds to a family (s − 1)s∈S of elements of I such that

∑s∈S(∂t/∂s)(s − 1) = 0 for all t ∈ T by

part(b) above. Thus there is an exact sequence ZG(T ) ∂2→ ZG(S) ∂1→ I where ∂2et =∑s∈S(∂t/∂s)es and

∂1es = s − 1 (et and es are basis elements of their respective groups). Now ∂1 is surjective because

31

the elements s − 1 generate I as a left ZG-ideal by Exercise I.2.1(b), where S is a set of generators for

G = F (S)/R, and so we have the desired exact sequence ZG(T ) ∂2→ ZG(S) ∂1→ I → 0.Note that this reproves the second part of Exercise II.5.3(d).

2.4(d): Since the augmentation ideal I of ZF is free (by Exercise IV.2.3(b)), we have a free resolution

0→ I → ZF ε→ Z→ 0 of Z over ZF . Taking R-coinvariants and noting that (ZF )R ∼= Z[F/R] = ZG byExercise II.2.1, we obtain a complex IR → ZG → Z → 0 whose homology is H∗R because H1(R,Z) ∼=Ker(IR → ZG) by the dimension-shifting technique; ZF is an H∗-acyclic module by Proposition III.6.1[1]since it is free as a ZR-module by Exercise I.8.2, and it maps onto Z with kernel I. Since I is free with

basis (s− 1)s∈S , I ∼= ZF (S) and we have the exact sequence 0→ H1Rϕ→ ZG(S) ∂1→ ZG→ Z→ 0, where

we note that taking coinvariants is a right-exact functor.Now we can map the standard (bar) resolution of Z over ZR to the aforementioned free resolution:

· · · // F2∂2 //

φ2

F1∂1 //

φ1

F0∂0 //

φ0

Z //

id

0

0 // Ii // ZF ε // Z // 0

where φn>1 = 0, φ1[r] = r − 1 = Dr, and φ0[ ] = 1. This is a commutative diagram becauseφ0∂1[r] = φ0(r[ ] − 1) = r − 1 = i(r − 1) = iφ1[r] and φ1∂2[r1|r2] = φ1(r1[r2] − [r1r2] + [r1]) =r1(r2 − 1)− (r1r2 − 1) + r1 − 1 = r1r2 − r1 − r1r2 + 1 + r1 − 1 = 0. By applying the coinvariants functorand noting that φ1|R = D|R and H1R = Ker(∂1)R/Im(∂2)R ∼= Rab, we have a commutative diagram

R

D|R// I

H1Rϕ// ZG(S)

where the vertical arrows are quotient maps. The composite FD→ I → ZG(S) is a derivation such that

s 7→ es because s1s2 7→ (s1− 1) + s1 · (s2− 1) 7→ es1 + s1es2 . Thus the map ϕ is given by r mod [R,R] 7→∑s∈S(∂r/∂s)es, and we have the desired exact sequence 0 → Rab

θ→ ZG(S) ∂→ ZG ε→ Z → 0 where

∂es = s− 1 and θ(r mod [R,R]) =∑s∈S(∂r/∂s)es.

Note that this reproves the first part of Exercise II.5.3(d).

3.1(a): Let 0→ A→ Eπ→ G→ 1 be an extension and let α : G′ → G be a group homomorphism, and

consider the pull-back (fiber-product) E ×G G′ = (e, g′) ∈ E × G′ | π(e) = α(g′). The kernel of thecanonical projection p : E ×G G′ → G′ corresponds to g′ = 0 ⇒ α(g′) = 0 = π(e) ⇒ Kerπ ∼= A, and

thus we have an extension 0 → A → E ×G G′p→ G′ → 1 which by definition fits into the commutative

diagram

0 // A // Eπ // G // 1

0 // A // E ×G G′p

//

OO

G′ //

α

OO

1

This extension is classified up to equivalence (by fitting into the above commutative diagram) becausegiven another such extension [corresponding to E′] of G′ by A, commutativity of the right-hand squareimplies there is a unique map φ : E′ → E ×G G′ by the universal property of the pull-back, and thisgives commutativity of the right-half of the diagram below:

0 // Ai1 // E

π // G // 1

0 // Ai2 // E ×G G′

p//

ϕ

OO

G′ //

α

OO

1

0 // Ai3 // E′ //

φ

OO

G′ //

XX

1Note that α for the E′-extension yields the identity map G′ → G′. It suffices to show that the left-side

32

of the diagram also commutes, for then we can apply the Five-Lemma which states φ is an isomorphism(E′ ∼= E ×G G′). Now φi3(a) = i2(b) for some b ∈ A because i3(a) maps to 0 ∈ G′ by exactness of thebottom row and hence lies in the kernel of E×GG′ which is contained in i2(A). Then ϕφi3(a) = ϕi2(b),and ϕφi3(a) = i1(a) by commutativity of the outer left-hand square while ϕi2(b) = i1(b) by commuta-tivity of the top left-hand square. Thus i1(a) = i1(b) ⇒ a = b by injectivity of the inclusion, and thisyields φi3(a) = i2(a) which gives commutativity of the bottom left-hand square and completes the proof.Therefore, α induces a map E(G,A) → E(G′, A), and this corresponds to H2(α,A) : H2(G,A) →H2(G′, A) under the bijection of Theorem IV.3.12[1].

3.1(b): Let 0 → Ai→ E

p→ G → 1 be an extension and let f : A → A′ be a G-module homomor-phism, and consider the largest quotient E′ of A′ o E such that the left-hand square in the followingdiagram commutes:

0 // Ai //

f

Ep//

φ

G // 1

0 // A′i′ // E′

p′// G // 1

Explicitly, E′ = A′oE/∼ with the equivalence relation (a′1, e1) ∼ (a′2, e2) iff a′1 +f(a1) = a′2 +f(a2) ande1 − i(a1) = e2 − i(a2) for some a1, a2 ∈ A; this relation is obviously reflexive and symmetric. It is tran-sitive because if a′1 +f(a1) = a′2 +f(a2) and a′2 +f(c) = a′3 +f(a3), then a′1 +f(a1) = a′3 +f(a3 +a2− c)and e1− i(a1) = [e3− i(a3) + i(c)]− i(a2) = e3− i(a3 + a2− c). Define i′ by i′(a′) = (a′, 0) and define p′

by p′(a′, e) = p(e) and define φ by φ(e) = (0, e). The map p′ is well-defined because for (a′1, e1) ∼ (a′2, e2)we have p′(a′2, e2) = p(e2) = p(e1) + p[i(a2 − a1)] = p(e1) + 0 = p(e1) = p′(a′1, e1). Now p′(a′, e) = 0 ⇒p(e) = 0 ⇒ ∃ a | i(a) = e ⇒ (a′, e) = (a′, i(a)) ∼ (a′ + f(a), 0) = i′(a′ + f(a)) ⇒ Kerp′ ⊆ Imi′, andp′[i′(a′)] = p′(a′, 0) = p(0) = 0 ⇒ Imi′ ⊆ Kerp′, so Kerp′ = Imi′ and the bottom row is an E′-extension.This extension is classified up to equivalence (by fitting into the above commutative diagram) becausegiven another such extension [corresponding to E′′] of G by A′, we get a diagram

0 // Ai //

f

Ep//

φ

G // 1

0 // A′i′ // E′

p′//

ϕ

G // 1

0 // A′i′′ // E′′

p′′// G // 1

where E → E′′ is the map Φ and the identity map A′ → A′ is induced from the f for the E′′-extension.We obtain an induced map ϕ : E′ → E′′ given by ϕ(a′, e) = i′′(a′) + Φ(e) which is well-defined be-cause if (a′1, e1) ∼ (a′2, e2) then ϕ(a′2, e2) = i′′(a′1) + i′′f(a1) − i′′f(a2) + Φ(e1) − Φi(a1) + Φi(a2) =[i′′(a′1) + Φ(e1)] + i′′f(a1) − i′′f(a2) − i′′f(a1) + i′′f(a2) = ϕ(a′2, e2) + 0 + 0 = ϕ(a′1, e1), noting thatΦi = i′′f by commutativity of the outer left-hand square. The bottom right-hand square is commutativebecause p′′ϕ(a′, e) = p′′i′′(a′) +p′′Φ(e) = 0 +p(e) = p(e) = p′(a′, e). The bottom left-hand square [hencethe whole diagram] is also commutative because ϕi′(a′) = ϕ(a′, 0) = i′′(a′) + Φ(0) = i′′(a′) + 0 = i′′(a′).We can now apply the Five-Lemma which states ϕ is an isomorphism (E′′ ∼= E′).Therefore, f induces a map E(G,A) → E(G,A′), and this corresponds to H2(G, f) : H2(G,A) →H2(G,A′) under the bijection of Theorem IV.3.12[1].

3.2(a): Let 0 → A′i→ A

p→ A′′ → 0 be a short exact sequence of G-modules and let d : G → A′′

be a derivation, and consider the set-theoretic pull-back E = (a, g) ∈ A o G | p(a) = d(g) where wenote that A o G = A × G as sets. If d : E → A and π : E → G are the canonical projections, thendπ[(a1, g1)(a2, g2)] = d(g1g2) = d(g1) + g1 · d(g2) = p(a1) + g1 · p(a2) and the group law on E can be thatof the semi-direct product due to the agreement pd(a1 +g1 ·a2, g1g2) = p(a1 +g1 ·a2) = p(a1)+g1 ·p(a2) =dπ(a1 + g1 · a2, g1g2). Thus E can be regarded as a subgroup of A o G, and d is a derivation becaused[(a1, g1)(a2, g2)] = d(a1 + g1 · a2, g1g2) = a1 + g1 · a2 = d(a1, g1) + g1 · d(a2, g2). Mimicking the proofof Exercise IV.3.1(a) verbatim, for each derivation d there is an extension 0 → A′ → E → G → 1characterized by the fact that it fits into a commutative diagram with derivation d

33

0 // A′i // A

p// A′′ // 0

0 // A′j// E

π //

d

OO

G //

d

OO

1This construction gives a map Der(G,A′′)→ E(G,A′).

3.2(b): What follows will be set-theoretic, and we use the same notation/maps as in part(a). A lifting ofd : G→ A′′ to a function l : G→ A is given by l(g) = p−1d(g), where p−1(x) = 0 if x /∈ Imp. This yieldsa cross-section s : G→ E of π given by s(g) = d−1l(g), because πs = πd−1l = πd−1p−1d = π(pd)−1d =π(dπ)−1d = ππ−1d−1d = idG. Consider the diagram

Der(G,A′′) //

E(G,A′)

∼=

H1(G,A′′)δ // H2(G,A′)

where the top-horizontal map is the constructed map of part(a) and the left-vertical map is the naturalquotient; the isomorphism (right-vertical map) sends an extension to a factor set f : G×G→ A′. Theconnecting homomorphism δ([d]) = [w2] is defined by lifting the element d of Z1(G,A′′) = Der(G,A′′)[equality comes from Exercise III.1.2] to an element w1 in C1(G,A) and then taking its image underthe coboundary map ∂∗w2 and noting that it must be the image of some w2 ∈ Z2(G,A′) under theinjection i∗ : C2(G,A′) → C2(G,A); all of this is well-defined. From above, w1 in this scenario is land so ∂∗l = L : G × G → A is defined by L(g, h) = gl(h) − l(gh) + l(g) via the equation on pg59[1].Then we have i∗(w2) = L, and w2 : G × G → A′ is a factor set which corresponds to a section S viaS(g)S(h) = j(w2(g, h))S(gh) by the normalization condition on pg91[1]. I claim that S is the sections defined above and that w2 is the factor set corresponding to s. Assuming this claim holds for themoment, the connecting homomorphism sends the [class of the] derivation d to the factor set w2 whichis precisely the image of the extension under the vertical-isomorphism, and that extension is the imageof d under the top-horizontal map, so commutativity of the diagram is satisfied. Therefore, it suffices toprove the claim, i.e. we must check the validity of the equation s(g)s(h) = j(w2(g, h))s(gh). Note thatj w2 = d−1L by commutativity of the left-side of the diagram in part(a), and s = d−1l, so the equationreduces to l(g) + [d−1l(g)] · l(h) = L(g, h) + [d−1L(g, h)] · l(gh) after we apply the derivation d to bothsides. The action of E on A is given by e · a = π(e)a, so [d−1l(g)] · l(h) = π[s(g)]l(h) = gl(h). Thus theequation becomes l(g)+gl(h) = [gl(h)− l(gh)+ l(g)]+[d−1L(g, h)] · l(gh) ⇒ l(gh) = [d−1L(g, h)] · l(gh).It now suffices to show that the element π[d−1L(g, h)] fixes l(gh), i.e. d−1L(g, h) ∈ A′ ⊂ E. But this isimmediate because d−1L(g, h) = j(w2(g, h)) ∈ A′ where A′ has lazily been identified with j(A′).

3.3: Let G be a finite group which acts trivially on Z. For any homomorphism G → Q/Z we canconstruct a central extension of G by Z by pulling back the canonical extension (the top row)

0 // Z // Q // Q/Z // 0

0 // Z // E //

OO

G //

OO

1Thus we have a map ϕ : Hom(G,Q/Z) → E(G,Z), which by Exercise IV.3.2 can be identified withthe boundary map δ : H1(G,Q/Z) → H2(G,Z) since H1(G,Q/Z) ∼= Hom(G,Q/Z) by Exercise III.1.2.Now H1G = Gab is finite, and so is H2G by Exercise AE.16 [G is finite and Z is finitely generated];thus Hom(H2G,Q) = 0 = Hom(H1G,Q) because there are no nontrivial homomorphisms from a finitegroup into the infinite group Q. Also, Q is Z-injective and hence Ext1

Z(H0G,Q) = 0 = Ext1Z(H1G,Q)

by Proposition 17.1.9[2]. The universal coefficient sequence of Exercise III.1.3 now yields H1(G,Q) =0 = H2(G,Q); alternatively, we could have simply noted that Hn(G,Q) = 0 ∀ n > 0 by CorollaryIII.10.2[1] since |G| is invertible in Q. By the long exact cohomology sequence (Proposition III.6.1[1]), δis an isomorphism. Thus ϕ is a bijection.

3.4(a): Let E be a group which contains a central subgroup C ⊆ Z(E) of finite index n.Method 1 : We have a central extension 1 → C → E → G → 1 by taking G = E/C. The nth

power map C → C (where C is considered as an abelian group) induces multiplication by n on H2(G,C)

34

because the cochain f : G2 → C is sent to nf : G2 → nC ⊆ C, and this is the zero-map because n = |G|annihilates H2(G,C) by Corollary III.10.2[1]. Now Exercise IV.3.1(b) implies the above central extensionfits uniquely into a commutative diagram with the split extension [the trivial element in E(G,C)]

1 // C //

n

E //

φ

G // 1

1 // C // C ×G // G // 1where we note that C o G ∼= C × G because C is central. Looking at the left-hand square with Erestricted to C, commutativity implies φ(c) = (cn, 1), so the C-component of E → C × G gives us thedesired homomorphism E → C whose restriction to C is the nth power map.

Method 2 : The abelianization map ρ : E → Eab composed with the transfer map tr : Eab → Cab = Cis given by ϕ : e 7→ e mod [E,E] 7→

∏g∈C\E ge(ge)

−1. Now |C\E| = n and c ∈ C commutes with all

elements of E, so ϕ(c) =∏g∈C\E gc(gc)

−1 = cn∏g(cg)−1 = cn

∏gg−1 = cn. Thus ϕ = tr ρ : E → C

is the desired homomorphism whose restriction to C is the nth power map.

3.4(b): Given a finitely generated group E, suppose the commutator subgroup [E,E] of E is finite.Then there are finitely many nontrivial elements g−1(ege−1) where g is a generator of E (and e ∈ Eis arbitrary), so there are only finitely many conjugates ege−1 of each generator g of E. Consider theinner automorphism group Inn(E) ∼= E/C, where C = Z(E) is the center of E. An arbitrary element isa function fe(x) = exe−1 [with e ∈ E fixed] which is determined by where it sends the generators of E.Since there are finitely many generators and finitely many conjugates of each generator, there are onlyfinitely many non-identity maps fe ∈ Inn(G). Thus |Inn(G)| is finite, and so the center C of E has finiteindex.Conversely, suppose the center C of the finitely generated group E has finite index. Then part(a) above

gives us a homomorphism ϕ : E → C such that ϕ|C is the nth power map Cn→ C. It suffices to show

that Kerϕ is finite, for then E/Kerϕ is abelian (being isomorphic to a subgroup of the abelian group Cby the First Isomorphism Theorem) and hence [E,E] ⊆ Kerϕ by Proposition 5.4.7[2], so the commutatorsubgroup [E,E] of E is finite. Now ϕ is the C-component of the map φ : E → C ×G given in part(a),and the kernel of φ is contained in the kernel of the projection π : E → G by commutativity of thatdiagram in part(a), so Kerϕ ⊆ Kerφ ⊆ Kerπ = C and hence Kerϕ ⊆ Ker(C

n→ C). It suffices to show

that C ⊆ E is a finitely generated abelian group, for then Ker(Cn→ C) is a subgroup of finite exponent

in a finitely generated abelian group (hence finite by the Fundamental Theorem of Finitely GeneratedAbelian Groups). The Nielsen-Schreier Theorem (Theorem 85.1[6]) states that every subgroup of a freegroup is free. The Schreier Index Formula (Theorem 85.3[6]) states that for a free group F of finiterank with a subgroup H of finite index, rkZH = |F : H|(rkZF − 1) + 1. The finitely generated groupE has the presentation F/R with F free of finite rank, and C ⊆ F corresponds bijectively to someH ⊆ F with H/R = C by the Lattice [4th] Isomorphism Theorem; H is free and finitely generatedby the Nielsen-Schreier Theorem and the Schreier index formula. Since quotients of finitely generatedgroups are finitely generated (the generators of the quotient are the images of the generators under theprojection), C is finitely generated (and abelian since it lies in the center of E).

3.5(a): Let E be a group which contains an infinite cyclic central subgroup of finite index.Method 1 : The group E of an extension in E(G,Z) gives us a homomorphism ϕ : G→ Q/Z by Exercise

IV.3.3 (where G = E/Z), along with a map φ : E → Q which is injective on Z ⊆ E. Now the nontrivialfinitely generated subgroups of Q are of the form m

n Z with m,n ∈ N, so Imφ = mn Z ∼= Z [note: E is finitely

generated because both Z and E/Z are]. Thus we have a surjective map φ : E → Imφ ∼= Z which is simplyφ rephrased. Its kernel F := Kerφ = Kerφ is finite because |F ∩Z| = 1 and |FZ/Z| = |F : F ∩Z| = |F | bythe 2nd Isomorphism Theorem, where we note that FZ/Z ⊆ G is finite [FZ is a subgroup of E because

F / E]. We then have an extension 0 → Kerφ → Eφ→ Z → 0 which must split (by Exercise AE.27)

because Z is free. Thus E ∼= F o Z with F finite.Method 2 : By Exercise IV.3.4(a) we have a map φ : E → Z which has a finite kernel as proved in Ex-

ercise IV.3.4(b). Letting φ : E → φ(E) be the surjective map induced from φ, we have φ(E) = mZ ∼= Zfor some m ∈ N and so φ : E → Z is a surjective map with finite kernel. We then have an extension0→ Kerφ→ E → Z→ 0 which must split (by Exercise AE.27) because Z is free. Thus E ∼= F oZ with

35

F finite (taking F = Kerφ = Kerφ).

3.5(b): Let E be a torsion-free group which has an infinite cyclic subgroup Z of finite index. LetE act by left multiplication on the finite set T of left cosets of Z in E and let πZ : E → S|T | be the asso-ciated permutation representation afforded by this action. By Theorem 4.2.3[2], the kernel of the actionis the core subgroup KerπZ =

⋂e∈E eZe−1 ≡ A and so A ⊆ Z is a normal subgroup of E of finite index

(A is necessarily nontrivial and hence infinite cyclic). The group action of E on A ∼= Z by conjugationis a homomorphism E → Aut(Z) = ±1, and the kernel is a subgroup E′ ⊆ E of index 1 or 2 suchthat A ⊆ Z(E′). By part(a) we have E′ ∼= F oA with F finite, but since E is torsion-free, F = 1 andhence E′ ∼= Z. We therefore have an extension 0→ Z→ E → G→ 1 with |G| ≤ 2 (take G = E/E′). IfG = 1 we are done. If G = Z2 acts non-trivially on Z then H2(G = Z2,Z) = ZZ2/NZ = 0/NZ = 0,where N ∈ ZG is the norm element. In view of Theorem IV.3.12[1] we see that the extension splits (soE ∼= Z oG), contradicting the assumption that E is torsion-free. Hence G acts trivially, so Z is centralin E and E ∼= Z by part(a).

3.6: Let E be a finitely generated torsion-free group which contains an abelian subgroup of finiteindex. This subgroup is isomorphic to Zn for some n by the Fundamental Theorem of Finitely Gen-erated Abelian Groups. Note that Zn is a polycyclic group because it is solvable (it has the series1 / Z / Zn with Zn/Z = Zn−1 abelian) and every subgroup is finitely generated; an equivalent defi-nition of a polycyclic group is that it has a subnormal series with each quotient cyclic (so for Zn wehave the series 1 / Z / Z2 / · · · / Zn−1 / Zn with each quotient Zi/Zi−1 = Z cyclic). Thus E is avirtually polycyclic group because it has a polycyclic subgroup of finite index [note: E is also virtu-ally abelian]. Let E act by left multiplication on the finite set T of left cosets of Zn in E and letπZn : E → S|T | be the associated permutation representation afforded by this action. By Theorem4.2.3[2], the kernel of the action is KerπZn =

⋂e∈E eZne−1 =: A and so A ⊆ Zn is a normal subgroup

of E of finite index; A is necessarily nontrivial and hence isomorphic to Zn because it is a subgroupof Zn of finite index (|E : Zn| = |(E/A)/(Zn/A)| is finite and |E : A| is finite). We therefore have an

extension 0 → Zn → Ep→ G → 1 with G finite. The group action of G on Zn is a homomorphism

ρ : G → Aut(Zn) ∼= GLn(Z), and GLn(Z) contains only integral matrices with determinant ±1 as de-duced from the surjective map det : GLn(Z) → Z× = ±1 or from the fact that an integral matrixis invertible iff its determinant is a unit in Z. Consider the finite kernel K := Kerρ and its preimageE′ := p−1(K) ⊆ E under p, and note that we have E′ /E by the Lattice Isomorphism Theorem becauseE′/Zn = K / G = E/Zn. Now this torsion-free group E′ is finitely generated because its subgroup Znand its quotient K are both finitely generated [if e ∈ E′−Zn then it can be written as a finite sum withgenerators of K, and if e ∈ Zn then it can be written as a finite sum with generators of Zn]. Also, thecorresponding K-action on Zn is trivial (as K is the kernel of the G-action) and hence Zn lies in thecenter of E′, so by Exercise IV.3.4(b) the commutator subgroup [E′, E′] is finite. But the only finitesubgroup of a torsion-free group is the trivial group, so [E′, E′] = 0 and E′ab = E′/[E′, E′] = E′ (i.e. E′

is abelian, hence isomorphic to Zn by the Fundamental Theorem of Finitely Generated Abelian Groups,noting that |E′ : Zn| = |K| is finite). Letting F = E/E′ ∼= (E/Zn)/(E′/Zn) = G/K where the isomor-phism follows from the 3rd Isomorphism Theorem, we have a group extension 0 → Zn → E → F → 1coupled with the faithful group action F → GLn(Z); this action is faithful because we modded the mapρ by its kernel and injected A ∼= Zn into E′ ∼= Zn.Since |F | = r annihilates H2(F,Zn) by Corollary III.10.2[1], the rth power map Zn → Zn induces thezero-map on H2(F,Zn). Exercise IV.3.1(b) then implies the above extension fits uniquely into a com-mutative diagram with the split extension [the trivial element in E(F,Zn)]

0 // Zn i1 //

r

Ep1 //

φ

F // 1

0 // Zn i2 // Zn o Fp2 // F // 1

The map φ is injective because if φ(e) = 0 then p1e = 0 ⇒ ∃ z | i1z = e ⇒ i2(rz) = φ(i1z) = 0 ⇒rz = 0 ⇒ z = 0 ⇒ e = i1z = i10 = 0. Alternatively, we could simply note that since the r-map isinjective, Kerφ∩Zn = 1 and hence Kerφ injects into F , which means Kerφ is trivial by commutativityof the right-side diagram.A crystallographic group is a discrete cocompact subgroup of the group Rn o On of isometries of some

36

Euclidean space; in general, V ⊆ W is cocompact if W/V is compact. Note that Rn has the usualtopology (its basis consists of the open n-balls), and the relative topology for the subspace Zn is thediscrete topology (any point x ∈ Zn is equal to the intersection Zn ∩ Bx where Bx is the ball of radius13 centered at x). The orthogonal group On = M ∈ GLn(R) | MTM = 1 has the relative topology

induced from the matrix group Mn(R) ∼= Rn2

.It is a fact that a faithful action F → GLn(R) can also be considered as a faithful action F → On,so in our case we have an injection F → On since GLn(Z) ⊂ GLn(R). The product topology onZnoF ⊂ RnoOn is the discrete topology [thus ZnoF is discrete] because Zn has the discrete topologyas mentioned above and the relative topology on F is discrete since F is finite. Since a subgroup of adiscrete group is discrete (the relative topology induced from the discrete topology is discrete), E canbe embedded (by φ) as a discrete subgroup of Rn oOn.The Zn-action on Rn given by translation x 7→ x + z is properly discontinuous, so the quotient mapp : Rn → Rn/Zn ∼= Tn is a regular covering map (see Exercise I.4.2) where Tn ≡ S1 × · · · × S1 is thecompact n-torus. Now On is a compact space by the Heine-Borel Theorem because it is a closed boundedsubspace of Rn2

(see pg292[3]), so Zn is a cocompact subgroup of Isom(Rn) := Rn o On because thequotient Tn o On is compact. But Zn is a subgroup of E of finite index, so E is also cocompact. Thiscompletes the proof that E is a crystallographic group.

3.7(a): Let G be a perfect group (so H1G = 0) and let A be an abelian group with trivial G-action.The universal coefficient sequence of Exercise III.1.3 then implies H2(G,A) ∼= Hom(H2G,A).

3.7(b): Let G be a perfect group and let A be any abelian group with trivial G-action. Yoneda’s lemmafrom Exercise I.7.3(a) states that a natural transformation ϕ : Hom(H2G,−)→ H2(G,−) is determinedby where it sends idH2G ∈ Hom(H2G,H2G), and so for the isomorphism ϕ of part(a) there is an elementu ∈ H2(G,H2G) such that ϕ(idH2G) = u. Now for any v ∈ H2(G,A) there is a unique map f : H2G→ Asuch that ϕ(f) = v because ϕ is an isomorphism, and Yoneda’s lemma gives ϕ(f) = H2(G, f)u. Thusu is the “universal” cohomology class of H2(G,H2G), in the sense that for any v ∈ H2(G,A) there is aunique map f : H2G→ A such that v = H2(G, f)u.

3.7(c): In view of Theorem IV.3.12[1], part(b) can be reinterpreted as saying that the [perfect] group

G admits a “universal central extension” 0→ H2G→ Eπ→ G→ 1 characterized by a certain property.

This property states that given any abelian group A and any central extension 0→ A→ E′π′→ G→ 1,

there is a unique map f : H2G→ A such that the extension is the image of the universal extension underH2(G, f) = E(G, f). By Exercise IV.3.1(b) this latter part means there is a map E → E′ making thefollowing diagram commute

Eπ //

G

E′π′

>>

and we assert that this map is unique. Indeed, two such maps h1, h2 : E ⇒ E′ which induce the samemap f must differ by a homomorphism φ : E → A because π′h1(e) = π(e) = π′h2(e) ⇒ h1e = h2e · awith a ∈ Kerπ′ = A, giving φ(e) = a [it is obviously a homomorphism since h1 and h2 are]; φ mustalso factor through G (i.e. φ is equal to E → E/H2G ∼= G → A) because h1 and h2 must agree onwhere it sends H2G by commutativity of the diagram in Exercise IV.3.1(b). Now any homomorphismψ : G→ A satisfies [G,G] ⊆ Kerψ by Proposition 5.4.7[2]; but G is perfect, so G = Kerψ and there areno non-trivial maps G→ A (hence φ = 0 ⇒ h1 = h2).

Note: It happens to be true that E in the universal central extension is necessarily perfect, so thereis another way to show uniqueness of the map E → E′ [this is presented in John Milnor’s Introduction toAlgebraic K-Theory ]. For any y, z ∈ E we have h1y = h2y·c and h1z = h2z ·c′ with c, c′ ∈ Kerπ′ ⊆ Z(E′).Thus h1(yzy−1z−1) = h2(yzy−1z−1) by basic rearrangements of the two previous equations, noting thatwe can move c and c′ around as they lie in the center of E′. Since E = [E,E], it is generated bycommutators and hence h1 = h2.

3.8(a): Let 0→ Ai→ E

π→ G→ 1 be a central extension with G abelian. The commutator pairing associ-ated to the extension is the map c : G×G→ A defined by c(g, h) = i−1([g, h]) = i−1(ghg−1h−1), where g

37

and h are lifts of g and h to E. Since A/E and G = E/A is abelian, [E,E] ⊆ A by Proposition 5.4.7[2] andso i−1([g, h]) is defined. Any other lift of g to E is of the form ga = ag, and [ag, a′h] = [g, h] because A liesin the center of E. Thus c is well-defined, and it is alternating because c(g, g) = i−1([g, g]) = i−1(1) = 0.Now gh is a lifting of g + h because π(gh) = π(g) + π(h) = g + h, and [gh, k] = ghkh−1g−1k−1 =g[h, k]g−1[g, k] = [g, k][h, k] where the last equality follows from [E,E] ⊆ A ⊆ Z(E). We then havec(g + h, k) = c(g, k) + c(h, k) because i is injective, giving i−1([g, k][h, k]) = i−1([g, k]) + i−1([h, k]). Ananalogous computation gives c(k, g + h) = c(k, g) + c(k, h), so c is Z-bilinear. Since c is alternating and

bilinear, c(g, h) = −c(h, g) and hence c can be viewed as a map∧2

G→ A, where∧2

G = G⊗G/〈g⊗g〉is the second exterior power of G.

3.8(b): Let f be a factor set to the central extension in part(a). To show that c(g, h) = f(g, h)− f(h, g)it suffices to show that [g, h] = i[f(g, h)]i[f(h, g)]−1 because i is injective. Given the section s : G→ E,f satisfies s(g)s(h) = i[f(g, h)]s(gh) and s(h)s(g) = i[f(h, g)]s(gh). Thus we have [s(g), s(h)] =s(g)s(h)s(g)−1s(h)−1 = i[f(g, h)]i[f(h, g)]−1. Since s(g) is a lifting of g for all g ∈ G, we have thedesired [s(g), s(h)] = [g, h].

3.8(c): Let θ : H2(G,A) → Hom(∧2

G,A) be the map which sends the class of a cocycle f to thealternating map f(g, h)−f(h, g). This map is well-defined because [δc](g, h)− [δc](h, g) = c(g) +gc(h)−c(gh) − c(h) − hc(g) + c(hg) = c(g) + c(h) − c(gh) − c(h) − c(g) + c(gh) = 0, where we note that G isabelian and the G-action on the cochain c is trivial since A is central. In view of Theorem IV.3.12[1],part(b) implies this image is the commutator pairing c(g, h), and f ∈ Kerθ iff θf = c(g, h) = 0. Nowc(g, h) = i−1([g, h]) = 0 iff [g, h] = 0 which is equivalent to E being abelian, i.e. [E,E] = 0. Thus thereis a bijection Kerθ ∼= Eab(G,A), where Eab(G,A) is the set of equivalence classes of abelian extensions ofG by A.

4.1: Let Q2n be a generalized quaternion group. It is a fact that Q2n has a unique element of or-der 2, hence a unique subgroup of order 2. Any subgroup of Q2n is also a 2-group (by Lagrange’sTheorem) and so it has an element of order 2 by Cauchy’s Theorem; this element is then unique in eachsubgroup (giving a unique Z2 subgroup). Therefore, by Theorem IV.4.3[1] every subgroup of Q2n iseither a cyclic group or a generalized quaternion group.Alternatively, Q2n = 〈x, y | x2n−1

= y4 = 1, yxy−1 = x−1〉 has the property that any subgroup G isa 2-group with a unique Z2 subgroup (as mentioned above). If G is abelian, then by the FundamentalTheorem of Finitely Generated Abelian Groups, G ∼= Z2α1 ×· · ·×Z2αi which has more than one Z2 sub-group if i > 1, and so G ∼= Z2α is a cyclic subgroup. Suppose G is nonabelian, which means G containselements of the form xk (they necessarily form a cyclic subgroup H ⊂ 〈x〉) and elements of the form xky[note: each element of Q2n can be written uniquely in the form xiyj for 0 ≤ i ≤ 2n−1− 1 and 0 ≤ j ≤ 1,and any element in Q2n − 〈x〉 has order 4 because (xky)4 = (xkyxky)2 = (xkx−kyy)2 = (y2)2 = y4 = 1].Let X = xk1y, xk2y ⊂ G be the elements of the form xky and let xi be the generator of H. Now

xk2y · xk1y = xk2−k1y2 = xMk+2n−2

and (xk1y)2 = y2, so xMky2 · y2 = xMk and xMk · xk1y = xk2y. Thusxk1y is a generator of G (of order 4) and the other generator is xr = xgcd(i,Mk) which is cyclic [note: if Xincludes other elements (up to xkmy) then the above still applies with r = gcd(i, km−k1, . . . , km−km−1),and if either i = 0 or k1 = · · · = km = 0 then omit those integers/differences in the gcd-term]. Since(xk1y)xr(xk1y)−1 = xk1(yxry−1)x−k1 = xk1x−rx−k1 = x−r, the presentation is complete and G is ageneralized quaternion group.

4.2: The dihedral group D2n = Z2n−1 o Z2 = 〈x, y | x2n−1

= y2 = 1, xy = yx−1〉 has the prop-erty that each element can be written uniquely in the form yixj for 0 ≤ i ≤ 1 and 0 ≤ j ≤ 2n−1 − 1.Any element not in Z2n−1 = 〈x〉 has order 2 because (yxk)2 = yxkyxk = (yxky)xk = x−kxk = 1.If a subgroup contains only elements of 〈x〉 then it is cyclic, so we consider the only other subgroupsG, and these contain elements of the form yxk and elements in 〈x〉 (which necessarily form a sub-group H ⊆ 〈x〉). Let X = yxk1 , . . . , yxkm ⊂ G be the elements of the form yxk and let xi be thegenerator of H. The given subset X implies G contains the elements xkm−k1 , . . . , xkm−km−1 . Thenxr = xgcd(i,km−k1,...,km−km−1) generates the elements in G of the form xk (because k would be a multipleof r), and yxk1 · x(km−k1)−(km−kj) = yxkj for any j ≤ m [note: if i = 0 (meaning the only elements ofG are of the form yxk) then r = gcd(km − k1, . . . , km − km−1), and if k1 = · · · = km = 0 (meaning y is

38

the only element in G of the form yxk) then r = i]. Thus G = 〈xr, yxk1〉 where yxk1 is of order 2 and(yxk1)xr(yxk1) = (yxk1+ry)xk1 = x−k1−rxk1 = x−r, i.e. G is a dihedral group.The non-cyclic abelian subgroups of D2n are isomorphic to Z2 × Z2 = D4. The group D8 contains twonon-cyclic abelian normal subgroups, 〈x2, y〉 and 〈x2, yx〉. We assert that such subgroups are not normalif n > 3 (i.e. if |D2n | > 8). From the subgroup presentations above we see that the non-cyclic abelian

subgroups are Hi = 〈x2n−2

, yxi〉 for all i ∈ 0, . . . , 2n−1 − 1. It suffices to show that there is an integerj such that the element xj(yxi)x−j = yxi−2j does not lie in Hi (i.e. i − 2j is neither i nor i + 2n−2

modulo 2n−1). The first condition implies 2j 6≡ 0 mod 2n−1 ⇒ j 6= 2n−2m for any m ∈ Z, and the lattercondition implies 2j + 2n−2 6≡ 0 mod 2n−1 ⇒ j 6= 2n−2m− 2n−3. Thus we can take j = 2n−2 − 1, andx2n−2−1 ∈ D2n will yield the non-normality (x2n−2−1)Hi(x

1−2n−2

) * Hi [note: if n = 3 then j cannot bean even integer 2m nor can it be an odd integer 2m− 1, which means j does not exist].

4.3: Let G = Zq o Z2 with q = 2n (n ≥ 3) and let A ⊂ Zq be the subgroup of order 2. Note thatA o Z2 = A × Z2 is a non-cyclic abelian subgroup of G since Z2 acts trivially on A [multiplication by−1 + 2n−1 is the action, and (−1 + 2n−1)2n−1 = −2n−1 + 2n−22n = 2n−1 + 0 = 2n−1 for the generator2n−1 ∈ A]. If H ⊂ G is a non-cyclic proper subgroup, then H * Zq and HZq is a subgroup of G byCorollary 3.2.15[2], so HZq = G. Thus 2q = |G| = |HZq| = |H|q/|H ∩ Zq| ⇒ |H : H ∩ Zq| = 2, andsince H ∩ Zq / H we have H/(H ∩ Zq) ∼= Z2 and this yields the extension 0→ H ∩ Zq → H → Z2 → 0.The generator of Z2 acts as multiplication by −1 on H ∩ Zq because −1 + 2n−1 ≡ −1 mod 2m withm < n [note: 2m = |H ∩ Zq| with m < n because H ∩ Zq is a subgroup of Zq and hence hasorder 2m, and if m = n then Zq ⊆ H which means H = G, contradicting the assumption thatH is proper]. If H were abelian then H ∩ Zq would be central in H, and if H ∩ Zq were cen-tral in H then H would be abelian since H/(H ∩ Zq) ∼= Z2 is cyclic; this latter fact holds becauseH/(H ∩ Zq) = 1(H ∩ Zq), x(H ∩ Zq) and so any h ∈ H has a representation h = xaz for z ∈ H ∩ Zq,giving h1h2 = xa1z1x

a2z2 = xa1xa2z1z2 = xa2xa1z2z1 = xa2z2xa1z1 = h2h1. Now H ∩Zq is central in H

iff the Z2-action is trivial (−z = z) iff H ∩ Zq = A, and so H is abelian iff H ∩ Zq = A.We have H1(Z2,Zq) = KerN where N : (Zq)Z2 → (Zq)Z2 is the norm map. But (Zq)Z2 is a quotientof a cyclic group (it is then necessarily cyclic of order m = 2r where m|q), and in particular we musthave (−1 + 2n−1)1 = 1 modm ⇒ 2n−2 − 1 = ms = 2(2r−1s) which is impossible because an evennumber cannot equal an odd number, so (Zq)Z2

= 0 and the kernel of the norm map is trivial. ThusH1(Z2,Zq) = 0, and by Proposition IV.2.3[1] this means that the extension 0→ Zq → G→ Z2 → 0 hasa unique splitting (up to conjugacy) and hence that G contains only two conjugacy classes of subgroupsof order 2 (specifically, 0× Z2 lies in one class, and A× 0 is the other class because the number of con-jugates of A is |G : NG(A)| = |G : G| = 1 by Proposition 4.3.6[2]). As an abelian non-cyclic subgroup,H contains at least two subgroups of order 2 (one necessarily being A) and hence H can be A× Z2 andits conjugates.We assert that H = A × Z2 is not normal in G. Indeed, the element (1, x) ∈ G where x is the genera-tor of Z2 can be used with (2n−1, 1) ∈ H to obtain (1, x)(2n−1, 1)(1, x)−1 = (1 + x · 2n−1, x)(−1, x) =(1 + 2n−1 + x · (−1), x2) = (2 + 2n−1, 1) which is not in H since 2 + 2n−1 is neither 0 nor 2n−1 modulo2n for n ≥ 3.

4.4: Let G be a p-group such that every abelian normal subgroup is cyclic and choose a maximal abeliannormal subgroup Zq ⊂ G with q = pn, so we have the corresponding extension 0→ Zq → G→ H → 1.If |H| = 1 then G ∼= Zq is cyclic, hence of type (A). If |H| = p then Theorem IV.4.1[1] states G is oftype (A),(D),(E), or (F) because groups of type (C) have a non-cyclic abelian subgroup of index p byProposition IV.4.4[1] (such subgroups are necessarily normal by Corollary 4.2.5[2]) and groups of type(B) are non-cyclic abelian normal subgroups of themselves [note: if G is of type (D) then we must have|G| ≥ 16 = 24 because G = D8 contains two non-cyclic abelian normal subgroups and G = D4

∼= Z2×Z2

is a non-cyclic abelian normal subgroup of itself]. Suppose, then, that |H| ≥ p2, and consider the normalsubgroups H ′ ⊆ H of order p. If such an H ′ acted trivially on Zq, then the inverse image G′ ⊂ Gof H ′ = G′/Zq would be a central extension of H ′ ∼= Zp by Zq, hence an abelian subgroup of G [asexplained in Exercise IV.4.3] bigger than Zq. But G′ / G by the Lattice Isomorphism Theorem (sinceH ′ / H), so it contradicts the maximality of Zq and hence H ′ cannot act trivially on Zq. Now G′ isa p-group with a cyclic subgroup Zq of index p, so by Theorem IV.4.1[1] it is of type (C),(D),(E), or(F) because types (A) and (B) were eliminated by the previous statement. Also, if H ′ acted as in (C)

39

then we would have a unique non-cyclic abelian subgroup G′′ of G′ of index p by Proposition IV.4.4[1];this applies to all p and n ≥ 2 except for the case p = 2 = n. Now G acts on G′ by conjugation (sinceG′ / G) and hence maps G′′ to another non-cyclic p-index subgroup. But G′′ is the only such subgroup,so conjugation sends G′′ to itself (i.e. G′′ / G) and G is a non-cyclic abelian normal subgroup of G,contradicting the hypothesis. Thus the only possibility is that p = 2 since groups of type (D),(E), and(F) are 2-groups, and the non-trivial element of H ′ acts as −1 [corresponding to (D),(E), and (C)] or−1 + 2n−1 with n ≥ 3 [corresponding to (F)]. This means that H ′ embeds in Z∗2n ∼= Z2 × Z2n−2 aseither Z2 × 0 or (0, 0), (1, 2n−3). But the composite H → Z∗2n → ±1 [where the first map is theaction-representation and the latter map is the projection] has a non-trivial kernel, and we can simplytake H ′ to be contained in the kernel; this implies H ′ is embedded as 0 × 2n−3Z/2n−2Z ⊂ Z∗2n whichis not any of the aforementioned subgroups. Thus we do not have |H| ≥ p2 and the proof is complete.

6.1: Suppose N is a group with trivial center.Method 1 : The center of N is C = 1, and so H2(G,C) = 0 = H3(G,C) for any group G. Then any

homomorphism ψ : G→ Out(N) gives rise to an obstruction in H3(G,C) = 0 which necessarily vanishes,thus E(G,N,ψ) 6= ∅ by Theorem IV.6.7[1]. Therefore, by Theorem IV.6.6[1], E(G,N,ψ) ∼= H2(G,C) = 0and hence there is exactly one extension of G by N (up to equivalence) corresponding to any homomor-phism G→ Out(N).

Method 2 : Note that N is an Aut(N)-crossed module via the canonical map α : N → Aut(N) andthe canonical action of Aut(N) on N , and Kerα = Z(N) = 1. Thus any extension of G by N fits intoa commutative diagram with exact rows

1 // Ni // E

π //

G //

ψ

1

1 // Nα // Aut(N) // Out(N) // 1

where ψ is determined by the E-extension. By Exercise IV.3.1(a), this is a pull-back diagram and henceall such extensions for a given ψ are equivalent (under the above diagram). Thus there is exactly oneextension of G by N (up to equivalence) corresponding to any homomorphism G→ Out(N).

6.2: Let 1 → N → E → G → 1 be an extension of finite groups such that gcd(|N |, |G|) = 1. Corol-lary IV.3.13[1] states that such an extension must split if N is abelian, and it is indeed true that thisresult could be generalized to the non-abelian case. Now one might hope to deduce this generalizationdirectly from Theorem IV.6.6[1] in view of the vanishing of H2(G,C) [note that H2(G,C) = 0 by thecohomology analogue of Exercise AE.16 because gcd(|C|, |G|) = 1, where C is the center of N ]. However,this does not work because E(G,N,ψ) doesn’t contain the semidirect product (hence the split extension)N o G unless ψ : G → Out(N) [which is determined by the above extension] lifts to a homomorphismϕ : G→ Aut(N).

40

5 Chapter V: Products

1.1: Let G = 〈t〉 be a finite cyclic group of order m and let F be the periodic resolution · · · → ZG t−1→ZG N→ ZG t−1→ ZG ε→ Z → 0, where N =

∑m−1i=0 ti is the norm element (note that Fn = ZG for all

n ≥ 0). Let ∆ : F → F ⊗ F be the map whose (p, q)-component ∆pq : Fp+q → Fp ⊗ Fq is given by

∆pq(1) =

1⊗ 1 p even

1⊗ t p odd, q even∑0≤i<j≤m−1 t

i ⊗ tj p odd, q odd

Now ∆ is a G-module map of degree 0, where ∆(g · x) = (g, g) · ∆(x) with the action given by re-striction of scalars with respect to the diagonal embedding G → G × G, so in order to prove that it isa diagonal approximation it suffices to show that it commutes with the boundary maps (making it achain map) and is augmentation-preserving. Note that (F ⊗ F )n =

⊕p+q=n Fp ⊗ Fq with differential

D(f ⊗f ′) = df ⊗f ′+(−1)degff ⊗df ′, where d is the boundary operator of F . In dimension zero we have(ε⊗ ε)[∆00(1)] = ε(1)⊗ ε(1) = 1⊗ 1 ≡ 1 = ε(1), so ∆ is augmentation-preserving [the equivalence equa-

tion is from Z⊗ZZ∼=→ Z]. Moving up a dimension, we have ∆00[(t−1)1] = (t, t)[1⊗1]−[1⊗1] = t⊗t−1⊗1

and D1[∆10(1) + ∆01(1)] = D1(1⊗ t) +D1(1⊗ 1) = (t− 1)⊗ t+ (−1)11⊗ 0 + 0⊗ 1 + (−1)01⊗ (t− 1) =t ⊗ t − 1 ⊗ t − 0 + 0 + 1 ⊗ t − 1 ⊗ 1 = t ⊗ t − 1 ⊗ 1, so commutativity is satisfied. Moving up anotherdimension, we have (∆10 + ∆01)[N1] =

∑k(tk, tk)[1⊗ t] +

∑k(tk, tk)[1⊗ 1] =

∑k tk ⊗ tk+1 +

∑k tk ⊗ tk

and D2[∆02(1) + ∆11(1) + ∆20(1)] = D2(1⊗ 1) +∑i<j D2(ti⊗ tj) +D2(1⊗ 1) = 0⊗ 1 + (−1)01⊗N1 +∑

i<j [(t−1)ti⊗ tj +(−1)1ti⊗(t−1)tj ]+N1⊗1+(−1)21⊗0 = 0+∑k 1⊗ tk+

∑i<j t

i+1⊗ tj−∑i<j t

i⊗tj −

∑i<j t

i ⊗ tj+1 +∑i<j t

i ⊗ tj +∑k tk ⊗ 1 + 0 =

∑k 1⊗ tk + (

∑k tk ⊗ tk − 1⊗ 1)− (

∑k 1⊗ tk − 1⊗

t) + (∑k tk ⊗ tk+1 − 1⊗ t− tm−1 ⊗ 1)− (

∑k tk ⊗ 1− 1⊗ 1− tm−1 ⊗ 1) =

∑k tk ⊗ tk+1 +

∑k tk ⊗ tk, so

commutativity is satisfied. Thus we can proceed by induction.For even integers p+q = 2c (c ∈ N) there are c+1 (1⊗1)-elements, 0 (1⊗t)-elements, and c (

∑i<j t

i⊗tj)-elements. For odd integers p+q = 2c+1 (c ∈ N) there are c+1 (1⊗1)-elements, 0 (

∑i<j t

i⊗tj)-elements,

and c+ 1 (1⊗ t)-elements. This then gives D2c[∆2c(1)] = c[N ⊗ 1 + 1⊗N ] + c∑i<j [t

i+1⊗ tj − ti⊗ tj+1]

and ∆2c−1(N1) = c∑k[tk ⊗ tk+1 + tk ⊗ tk], and these are equal because they are just a multi-

ple of the low-dimensional case; the c[N ⊗ 1 + 1 ⊗ N ] came from (c + 1)[N ⊗ 1 + 1 ⊗ N ] minusN ⊗ 1 + 1⊗N which accounts for the two zero-dimensional tensor components which give trivial bound-ary, i.e. D(1⊗ 1 + 1⊗ 1) = (0⊗ 1 +N ⊗ 1) + (N ⊗ 1 + 1⊗ 0). Commutativity in the next dimension isalso satisfied [similar calculation], and the proof is complete.

2.1: Let M (resp. M ′) be an arbitrary G-module (resp. G′-module), let F (resp. F ′) be a projective reso-lution of Z over ZG (resp. ZG′), and consider the map (F⊗GM)⊗(F ′⊗G′M ′)→ (F⊗F ′)⊗G×G′ (M⊗M ′)given by (x ⊗ m) ⊗ (x′ ⊗ m′) 7→ (x ⊗ x′) ⊗ (m ⊗ m′). Note that (F ⊗G M) ⊗ (F ′ ⊗G′ M ′) =(F ⊗ M)G ⊗ (F ′ ⊗ M ′)G′ , which is the quotient of F ⊗ M ⊗ F ′ ⊗ M ′ by the subgroup generated

by elements of the form gx⊗ gm⊗ g′x′ ⊗ g′m′. The isomorphism F ⊗M∼=→M ⊗ F of chain complexes

(M in dimension 0) is given by x ⊗ m 7→ (−1)degm·degxm ⊗ x = m ⊗ x, and so the aforementionedquotient is isomorphic to F ⊗ F ′ ⊗M ⊗M ′ modulo the subgroup generated by elements of the form(gx⊗g′x′⊗gm⊗g′m′) = (g, g′) ·(x⊗x′⊗m⊗m′) where this latter action is the diagonal (G×G′)-action.Now this is precisely (F ⊗ F ′ ⊗M ⊗M ′)G×G′ = (F ⊗ F ′) ⊗G×G′ (M ⊗M ′) and hence the consideredmap is an isomorphism.Assuming now that either M or M ′ is Z-free, we have a corresponding Kunneth formula⊕n

p=0Hp(G,M)⊗Hn−p(G′,M ′) → Hn(G×G′,M⊗M ′)

⊕n−1p=0 TorZ1 (Hp(G,M), Hn−p−1(G′,M ′))

by Proposition I.0.8[2]. Note that in order to apply the proposition we needed one of the chain complexes(say, F ⊗GM) to be dimension-wise Z-free (and so with a free resolution F this means we needed M tobe Z-free). Actually, the general Kunneth theorem has a more relaxed condition and it suffices to chooseM (or M ′) as a Z-torsion-free module.

*Must require conditions on groups/actions (say, one action is trivial?).

2.2: [[no proofs, just notes]] (for more info, see topological analog in §60[4])Let M (resp. M ′) be an arbitrary G-module (resp. G′-module), let F (resp. F ′) be a projective resolu-

41

tion of Z over ZG (resp. ZG′), and consider the cochain cross-product HomG(F,M)⊗HomG′(F′,M ′)→

HomG×G′(F ⊗ F ′,M ⊗M ′) which maps the cochains u and u′ to u × u′ given by 〈u × u′, x ⊗ x′〉 =(−1)degu′·degx〈u, x〉⊗〈u′, x′〉. This map is an isomorphism under the hypothesis that either H∗(G,M) orH∗(G

′,M ′) is of finite type, that is, the ith-homology group is finitely generated for all i (alternatively,we could simply require the projective resolution F or F ′ to be finitely generated). For example, ifM = Z then HomG(−,Z) commutes with finite direct sums, so we need only consider the case F = ZG.An inverse to the above map is given by t 7→ ε⊗ φ, where ε is the augmentation map and φ : F ′ →M ′

is given by φ(f ′) = t(1⊗ f ′). Note that this does not hold for infinitely generated P =⊕∞ ZG because

HomG(P,Z) ∼=∏∞ Z is not Z-projective (i.e. free abelian).

Assuming now that either M or M ′ is Z-free, we have a corresponding Kunneth formula⊕np=0H

p(G,M)⊗Hn−p(G′,M ′) → Hn(G×G′,M⊗M ′)⊕n+1

p=0 TorZ1 (Hp(G,M), Hn−p+1(G′,M ′))by Proposition I.0.8[2].

*Must require conditions on groups/actions (say, one action is trivial?).

3.1: Let m ∈ H0(G,M) = MG and u ∈ Hq(G,N), and let fm : H∗(G,N) → H∗(G,M ⊗ N) de-note the map induced by the coefficient homomorphism n 7→ m ⊗ n. This homomorphism is also given

by n 7→ 1⊗n 7→ m⊗n where the former map in the composition is the canonical isomorphism N∼=→ Z⊗N ,

the latter map in the composition is Fm⊗idN : Z⊗N →M⊗N , and Fm : Z→M is given by F (1) = m.Then using two properties of the cup product (existence of identity element and naturality with respectto coefficient homomorphisms) we obtain fm(u) = (Fm ⊗ idN )∗(1 ` u) = F ∗m(1) ` id∗N (u) = m ` u.Now let m ∈ H0(G,M) = MG and z ∈ Hq(G,N), and let fm : H∗(G,N) → H∗(G,M ⊗ N) denotethe map induced by the same coefficient homomorphism n 7→ m ⊗ n. Then using two properties of thecap product (existence of identity element and naturality with respect to coefficient homomorphisms)we obtain fm(z) = (Fm ⊗ idN )∗(1 a z) = F ∗m(1) a id∗N (z) = m a z.

3.2: Consider the diagonal transformation ∆ presented in Exercise V.1.1, along with the cohomol-ogy groups H2r(G,M) ∼= MG/NM and H2r+1(G,M ′) ∼= Ker(N : M ′ → M ′)/IM ′ of the finite cyclicgroup G = 〈σ〉 of order n, where I = 〈σ − 1〉 is the augmentation ideal of G. The cup product inH∗(G,M⊗M ′) is given by u ` v = (u×v)∆ with 〈u×v, x⊗x′〉 = (−1)degv·degx〈u, x〉⊗〈v, x′〉. Chooserepresentatives 〈u, x〉 = m ∈ M of Hi(G,M) with m ∈ MG for i even and m ∈ Ker(N : M → M)for i odd, and choose representatives 〈v, x′〉 = m′ ∈ M ′ of Hj(G,M ′) with m′ ∈ M ′G for j even andm′ ∈ Ker(N : M ′ →M ′) for j odd. If i is even then (−1)degv·degx = (−1)degv·i = 1, and if j is even then(−1)degv·degx = (−1)−j·degx = 1, and if both i and j are odd then (−1)degv·degx = (−1)−j·i = −1. Thusthe cup product element of Hi+j(G,M ⊗M ′) is represented by m⊗m′ for i or j even and is representedby −

∑0≤p<q≤n−1 t

pm⊗ tqm′ when i and j are both odd.

3.3(a): Let G be a finite group which acts freely on S2k−1, and consider the exact sequence of G-modules from pg20[1], 0 → Z → C2k−1 → · · · → C1 → C0 → Z → 0 where each Ci = Ci(S

2k−1) isfree. Tensoring the sequence with an arbitrary G-module M gives an exact sequence (as explained on

pg61[1]) 0 → M → C2k−1 ⊗M∂2k−1→ · · · → C1 ⊗M

∂1→ C0 ⊗M∂0→ M → 0. We can break this up into

short exact sequences 0 → Ker∂0 → C0 ⊗M → M → 0 and 0 → Ker∂i → Ci ⊗M → Ker∂i−1 → 0for all i, and we can then apply the H∗(G,−) functor to obtain corresponding long exact cohomol-ogy sequences. First note that HomG(F,C∗ ⊗M) → HomG(F,M) is a weak equivalence by TheoremI.8.5[1], so Hn(G,Ci ⊗ M) ∼= Hn(HomG(F,M)) = 0 for all i with n > 0 (M is considered a chaincomplex concentrated in dimension 0). Thus we can apply the dimension-shifting argument to the aboveshort exact sequences and obtain Hi(G,M) ∼= Hi+1(G,Ker∂0) ∼= Hi+2(G,Ker∂1) ∼= · · · ∼= Hi+2k(G,M)for i > 0 where we note that Ker∂2k−1 = M . For i = 0 we use the first short exact sequence men-tioned above to obtain the exact sequence H0(G,M) → H1(G,Ker∂0) → H1(G,C0 ⊗M) = 0, withH1(G,Ker∂0) ∼= H2k(G,M). Thus there is an iterated coboundary map d : Hi(G,M) → Hi+2k(G,M)which is an isomorphism for i > 0 and an epimorphism for i = 0.

3.3(b): Consider the “periodicity map” d from part(a) above with the finite group G which acts freelyon the sphere S2k−1. Since d is simply an iteration of coboundary maps δ, we can use Property V.3.3[1]which states d(w ` v) = d(w) ` v for any w ∈ H∗(G,Z) and v ∈ H∗(G,M). Choosing w = 1 ∈ H0(G,Z)and noting that 1 ` v = v by Property V.3.4[1], d(v) = d(1 ` v) = d(1) ` v. But d is an isomorphism

42

from H0(G,Z) to H2k(G,Z) by part(a) above, so there exists an element u ∈ H2k(G,Z) satisfyingd(1) = u. Thus there is an element u ∈ H2k(G,Z) such that the “periodicity map” d of H∗(G,M) isgiven by d(v) = u ` v for all v ∈ H∗(G,M).

3.3(c): Let G be a finite cyclic group of order |G| = n, and note that it acts freely on the circleS1 by rotations. By part(b) above, the periodicity isomorphism d maps 1 ∈ H0(G,Z) ∼= Z to a generatorα ∈ H2(G,Z) ∼= Zn. For the ring structure on H∗(G,Z) with multiplication being the cup product,α2 is a generator of H4(G,Z) ∼= Zn since d is an isomorphism and d(α) = α ` α = α2. General-izing, αm is a generator of H2m(G,Z) ∼= Zn and hence the cohomology ring is the polynomial ringH∗(G,Z) ∼= Z[α]/(nα) with |α| = 2.Alternatively, we could note that the infinite-dimensional lens space L∞ is a K(G, 1)-complex, soH∗(G,Z) ∼= H∗(L∞) by the cohomological analog of Proposition II.4.1[1]. The ring structure wascalculated in Example 3.41 on pg251[3], giving H∗(L∞) ∼= Z[α]/(nα) with |α| = 2.

3.4: Let G be cyclic of order n and consider the endomorpism α(m) of G given by α(m)g = gm,for any m ∈ Zn. Since α(m)∗(u ` v) = α(m)∗u ` α(m)∗v and H∗(G,Z) consists of elements of the form∑i ziβ

i with zi ∈ Z and |β| = 2 by Exercise V.3.3(c), it suffices to calculate α(m)∗ on H2(G,Z) in orderto calculate α(m)∗ : H∗(G,Z)→ H∗(G,Z). By Exercise III.1.3 we have the universal coefficient isomor-phism H2(G,Z) ∼= Ext(H1G,Z) since Hom(H1G,Z) = Hom(Zn,Z) = 0. The map α(m) is multiplicationby m on Gab = G and hence is multiplication by m on H1G by Exercise II.6.3(a). Now f(mg) = mf(g)for any group homomorphism f , so given a free resolution F of the abelian group Gab = H1G = G(namely, its presentation) we have the commutative diagram

0 Hom(F1,Z)oo

m

Hom(F0,Z)φoo

m

Hom(G,Z)oo

m

0 Hom(F1,Z)oo Hom(F0,Z)φoo Hom(G,Z)oo

Thus α(m) induces multiplication by m on Cokerφ ≡ Ext(H1G,Z) and hence α(m)∗ is multiplication bym on H2(G,Z) by naturality of the universal coefficient formula. Alternatively, we could have used theinterpretation of H2 in terms of group extensions (Exercise IV.3.1(a)) which states that α(m) inducesa map E(G,Z) → E(G,Z), sending the extension E to the fiber-product E ×G G which correponds tomE (i.e. the elements em ∈ E) and hence gives the m-multiplication map in cohomology by TheoremIV.3.12[1]. Referring back to the cohomology ring, α(m)∗ is multiplication by mi for the (2i)th-dimensionsince

∑i ziβ

i 7→∑i ziα(m)∗(β ` · · · ` β) =

∑i zi[α(m)∗β] ` · · · ` [α(m)∗β] =

∑i zi(mβ) ` · · · `

(mβ) =∑i zim

iβi.

3.5: The symmetric group S3 on three letters has a semi-direct product representation S3 = Z3 o Z2

where Z2 acts on Z3 by conjugation. Thus H∗(S3) = H∗(S3)(2)⊕H∗(S3)(3)∼= H∗(S3)(2)⊕H∗(Z3)Z2 by

Theorem III.10.3[1], with H∗(S3)(2) isomorphic to the set of S3-invariant elements of H∗(Z2). Exercise

III.10.1 showed that H∗(S3)(2)∼= Z2, so it suffices to compute H∗(Z3)Z2 where we know that Z2 acts by

conjugation on Z3 (1 7→ 1, x 7→ x2, x2 7→ x). But this action can be considered as the endomorphismα(2) from Exercise V.3.4 above since (1)2 = 1 and (x)2 = x2 and (x2)2 = x4 = x, and that exerciseimplies that the induced map on the (2i)th-cohomology is multiplication by 2i [we know that the co-homology is trivial in odd dimensions]. Now 21 ≡ 2 mod3 and 22 ≡ 1 mod3, so by multiplying both ofthose statements by 2 repeatedly we see that 2i ≡ 1 mod3 for i even and 2i ≡ 2 mod3 for i odd. Thusthe largest Z2-submodule of H2i(Z3) ∼= Z3 on which Z2 acts trivially is Z3 (itself) for i even, and is 0for i odd. It now follows that the integral cohomology H∗(S3) is the same as that which was deduced inExercise III.10.1, namely, it is Z2 in the 2 mod4 dimensions and is Z6 in the 0 mod4 dimensions and is 0otherwise (besides the 0th-dimension in which it is Z).

4.1:

4.2:

5.1(a): Let G be an abelian group so that ZG is a commutative ring, and let M ⊗ZG N be the tensor

43

product with ZG-module structure defined by r ·(m⊗n) = rm⊗n = m⊗rn, where r ∈ ZG. There exists

a Pontryagin product given by H∗(G,M)⊗H∗(G,N)×→ H∗(G×G,M ⊗N)

µ∗→ H∗(G,M ⊗ZG N). Theformer map is the homology cross product (see pg109[1]), where G×G acts on M⊗N via (g, g′)·(m⊗n) =gm⊗g′n. The latter map in the composition is induced by µ = (α, f) : (G×G,M⊗N)→ (G,M⊗ZGN),where α(g, g′) = gg′ and f(m ⊗ n) = m ⊗ n. As explained on pg79[1], µ∗ is a well-defined map as longas f(rm) = α(r)f(m) for r ∈ ZG. This is indeed the case because α(g, g′)f(m ⊗ n) = gg′ · (m ⊗ n) =g · (m⊗ g′n) = gm⊗ g′n = f(gm⊗ g′n) = f [(g, g′) · (m⊗ n)].

5.1(b): Take G = Z3 = 〈g〉 and take the modules M = N = Z7 with nontrivial Z3-action, i.e. g ·xi = x2i

and g2 · xi = x4i with g3 = 1 [consistency check: g3 · xi = x8i = (x8)i = xi]. Note that there is only onenontrivial action in this scenario because an action is a homomorphism Z3 → Aut(Z7) ∼= Z6 and thereare two such maps, the trivial map and the inclusion. The tensor product Z7 ⊗Z3

Z7 can be interpretedas the group (Z7 ⊗ Z7)Z3 where Z3 acts diagonally, g · (m ⊗ n) = gm ⊗ gn. This group contains onlyidempotent elements since (xi ⊗ xj)2 = x2i ⊗ x2j = gxi ⊗ gxj = g · (xi ⊗ xj) ∼ xi ⊗ xj . Now thetensor product Z7 ⊗Z[Z3] Z7 can be interpreted as the group (Z7 ⊗ Z7)Z3

where Z3 acts anti-diagonally,g · (m⊗ n) = g−1m⊗ gn = g2m⊗ gn. This group, however, does contain elements which are not idem-potent. For example, (x⊗ x)2 = x2 ⊗ x2 = g2x4 ⊗ gx ∼ x4 ⊗ x = (x⊗ x)(x3 ⊗ 1) and (x3 ⊗ 1) 1⊗ 1.Thus Z7 ⊗Z[Z3] Z7 Z7 ⊗Z3 Z7.

5.2: If G and G′ are abelian and k is a commutative ring, then the cohomology cross-product H∗(G, k)⊗kH∗(G′, k)

×→ H∗(G × G′, k ⊗ k) is a k-algebra homomorphism. Moreover, z × z′ = p∗z ` p′∗z′ for any

z ∈ H∗(G, k) and z′ ∈ H∗(G′, k), where p : G×G′ → G and p′ : G×G′ → G′ are the projections.To prove these two statements, note first that the cross-product is given by 〈z×z′, x⊗x′〉 = (−1)|z

′||x|z(x)⊗z′(x′) for x ∈ F and x′ ∈ F ′, with F and F ′ being the resolutions for G and G′, respectively. Then

〈(z1 ⊗ z′1)(z2 ⊗ z′2), x⊗ x′〉 = 〈(−1)|z′1||z2|z1z2 × z′1z′2, x⊗ x′〉 = (−1)|z

′1||z2|(−1)|z

′1z′2||x|z1(x)z2(x)⊗

z′1(x′)z′2(x′) = (−1)|z′1||x|(−1)|z

′1||x|[z1(x)⊗ z′1(x′)][z2(x)⊗ z′2(x′)] = 〈z1 × z′1, x⊗ x′〉 · 〈z2 × z′2, x⊗ x′〉

and so the cross-product is a k-algebra homomorphism (it obviously commutes with addition). Sincez ⊗ z′ = (z ⊗ 1) · (1 ⊗ z′) with 1 ∈ H0(G,Z) = Z, it follows that z × z′ = (z × 1) ` (1 × z′), where wemake the identifications Z⊗ k = k = k⊗Z to switch from tensor-multiplication to cup product. I claimthat z × 1 = p∗z; indeed, by naturality of the cross-product it suffices to check this for G′ = 1. Butthis is obvious because p∗ : H∗(G, k) → H∗(G × 1, k) maps z to z × 1, noting that F ′ = 1. Similarly,1× z′ = p′∗z

′, whence the proposition.

5.3(a): A directed set D is a partially-ordered set having the property that for each pair α, β ∈ Dthere exists γ ∈ D such that α, β ≤ γ. A directed system of groups is a family of groups Gαindexed by a directed set D along with a family of homomorphisms fαβ : Gα → Gβ such thatfαα = idGα and fαγ = fβγ fαβ for α ≤ β ≤ γ. The direct limit G = lim−→Gα of this directed systemis defined to be

∐αGα/∼ where gα ∼ gβ if there exists some γ ∈ D such that fαγ(gα) = fβγ(gβ).

Now for any G-module M , we have a compatible family of maps H∗(Gα,M) → H∗(G,M), hencea map ϕ : lim−→H∗(Gα,M) → H∗(G,M). It is true that homology commutes with direct limits ofchain complexes (see Albrecht Dold’s Lectures on Algebraic Topology, Proposition VIII.5.20), so toprove that ϕ is an isomorphism it suffices to show that lim−→(Fα ⊗Gα M) = F ⊗G M where Fα is thestandard resolution for Gα and F is the standard resolution for G. The obvious maps Fα → Fβare given by (g1, . . . , gn) 7→ (fαβ(g1), . . . , fαβ(gn)), and F is obviously the direct limit of Fα. ThusF ⊗GM = (

∐α Fα/∼) ⊗GM , and since we can switch actions from G to Gα via restriction of scalars,

(∐α Fα/∼)⊗GM =

∐α(Fα⊗GαM)/∼ where ∼ is simply altered by tensoring each tuple with an element

of M . But∐α(Fα⊗GαM)/∼ is by definition the direct limit of Fα⊗GαM , so lim−→(Fα⊗GαM) = F ⊗GM

and hence group homology commutes with direct limits of groups.

5.3(b): It is a fact that any group is the direct limit of its finitely generated subgroups. So for any abeliangroup G = lim−→Gα and commutative ring k, the homology ring H∗(G, k) is isomorphic to lim−→H∗(Gα, k)by part(a) above. If each of those rings H∗(Gα, k) is strictly anti-commutative, then H∗(G, k) will obvi-ously be strictly anti-commutative since it is a quotient of the direct sum of those rings. Thus we canreduce to the case where G is a finitely generated abelian group, hence isomorphic to a finite product

44

of cyclic groups. Let’s argue by induction on the number of cyclic factors. The infinite cyclic group hasresolution F =

∧(x) which is strictly anti-commutative as explained in subsection 5.2 on pg118[1], and

the finite cyclic group has resolution F =∧

(e1)⊗ZGΓ(e2) which is strictly anti-commutative by property(iii) of subsection 5.3 on pg119[1]. Since the admissible product on F induces a k-bilinear product onF ⊗G k via (f⊗k)(f ′⊗k′) = ff ′⊗kk′, the complex F ⊗G k is strictly anti-commutative; thus H∗(G, k) isstrictly anti-commutative for G cyclic. Applying the inductive hypothesis, we can attach another cyclicfactor by the method of subsection 5.4 on pg119[1]: F and F ′ are resolutions with admissible productfor G (cyclic) and G′ (inductive group), so F ⊗ F ′ is a resolution with admissible product for G × G′.This resolution is strictly anti-commutative because (x ⊗ y)(x ⊗ y) = (−1)|x|·|y|x2 ⊗ y2 = 0 if either x2

or y2 is 0, so for x (resp. y) of odd degree and y (resp. x) of even degree we have x ⊗ y of odd degreewhich satisfies (x⊗y)2 = 0. Then (F ⊗F ′)⊗G×G′ k is strictly anti-commutative and so is H∗(G×G′, k),completing the inductive process. Thus the ring H∗(G, k) is strictly anti-commutative for any abeliangroup G and commutative ring k.

5.4: Let n = p + q + r, and let σ ∈ Sn be a permutation with signature sgn(σ) being the numberof inversions of σ; an inversion of σ is a pair of elements (i, j) such that i < j and σ(i) > σ(j) [italso indicates the number of swaps needed to give the original sequence ordering]. A permutation σis called a (p, q, r)-shuffle if σ(i) < σ(j) for 1 ≤ i < j ≤ p and for p + 1 ≤ i < j ≤ p + q and forp + q + 1 ≤ i < j ≤ p + q + r. But this permutation is clearly the composition of a (p, q)-shuffle τ1and a (p + q, r)-shuffle τ2 since the former shuffle will give τ1(i) < τ1(j) for 1 ≤ i < j ≤ p and forp+ 1 ≤ i < j ≤ p+ q, and the latter shuffle will give τ2(i) < τ2(j) for p+ q + 1 ≤ i < j ≤ p+ q + r andwill preserve the ordering of the former shuffle via 1 ≤ i < j ≤ p + q. Then sgn(σ) = sgn(τ1) + sgn(τ2)because the inversions of τ2 give the original ordering of the sequence up to changes in 1, . . . , p+ q andthe inversions of τ1 give the original ordering of that set. Therefore (in the bar resolution),

[g1| · · · |gp] · [gp+1| · · · |gp+q] · [gp+q+1| · · · |gp+q+r] =∑σ σ[g1| · · · |gp+q+r]

where σ ranges over the (p, q, r)-shuffles.The notation is [g1| · · · |gn] · [gn+1| · · · |gn+m] =

∑τ τ [g1| · · · |gn+m] where τ ranges over all (p, q)-shuffles,

and σ[g1| · · · |gn] = (−1)sgnσ[gσ−1(1)| · · · |gσ−1(n)].Generalizing, we have

[g1| · · · |ga1 ] · [ga1+1| · · · |ga1+a2 ] · · · [ga1+···+an−1+1| · · · |ga1+···+an ] =∑σ σ[g1| · · · |ga1+···+an ]

where σ ranges over the (a1, . . . , an)-shuffles.

6.1: Let G be an abelian group (written additively) with n ∈ Z, and consider the endomorphism g 7→ ngof G. To see what it induces on the rational homology ring H∗(G,Q) it suffices to figure out what theendomorphism induces on the exterior algebra

∧∗(G ⊗ Q) by Theorem V.6.4[1] (they’re isomorphic),

noting that the isomorphism in said theorem is natural. Now the induced map on∧∗

(G⊗Q) is uniquely

determined by the induced map ϕ : G⊗Q→∧1

(G⊗Q) = G⊗Q by the universal mapping property ofexterior algebras (pg122[1]), and this map ϕ is given by g ⊗ q 7→ ng ⊗ q = n(g ⊗ q), i.e. multiplicationby n. Then on the i-fold tensor product of G ⊗ Q with itself (hence the exterior algebra) we have theinduced map as f1 ⊗ · · · ⊗ fi 7→ nf1 ⊗ · · · ⊗ nfi = ni(f1 ⊗ · · · ⊗ fi) where f1, . . . , fi ∈ G⊗Q. Thus theoriginal endomorphism on G induces multiplication by ni on H∗(G,Q) for the ith-dimension.

6.2: Let A and B be strictly anti-commutative graded k-algebras, where k is a commutative ring. We as-sert that A⊗kB is the sum (i.e. coproduct) of A and B in the category of strictly anti-commutative gradedk-algebras, via the maps fA : A→ A⊗kB and fB : B → A⊗kB with a 7→ a⊗1 and b 7→ 1⊗b. The coprod-uct refers to the pair (A⊗kB, fA, fB) satisfying the universal property that given a family of algebra ho-momorphisms gA : A→ C, gB : B → C, there exists a unique algebra homomorphism h : A⊗kB → Csuch that hfA = gA and hfB = gB . We define a k-bilinear map A×B → C given by a×b 7→ gA(a)gB(b).Then Corollary 10.4.16[2] gives us a unique homomorphism h : A⊗kB → C given by a⊗b 7→ gA(a)gB(b),and this is clearly an algebra homomorphism since h[(a1 ⊗ b1)(a2 ⊗ b2)] = h[(−1)|a2||b1|a1a2 ⊗ b1b2] =(−1)|a2||b1|gA(a1a2)gB(b1b2) = (−1)|a2||b1|gA(a1)[(−1)|a2||b1|gB(b1)gA(a2)]gB(b2) = h(a1 ⊗ b1)h(a2 ⊗ b2).Now hfA(a) = h(a ⊗ 1) = gA(a)gB(1) = gA(a), with a similar calculation for B, and so the universalproperty of tensor products (Theorem 10.4.10[2]) guarantees that A⊗k B is the categorical sum A+B.

45

6.3(a): Let A be a strictly anti-commutative graded ring with a differential ∂ and a system of di-vided powers, so that there is a family of functions ϕi : A2n → A2ni denoted x 7→ x(i) satisfying theproperties on pg124[1]. Note that x(i) is a cycle if x is a cycle because ∂x(i) = x(i−1)∂x = x(i−1) · 0 = 0.Thus ϕi restricted to the kernel Z2n ⊆ A2n gives a function Z2n → Z2ni. These functions inherit thesame properties associated with A because every term in all of the properties are cycles xi and x(j),assuming x is a cycle. Now suppose x(i) is a boundary whenever x is a boundary. Then ϕi induces afunction H2nA → H2niA because x = x + ∂y 7→ x(i) + (∂y)(i) = x(i) + ∂w = x(i), and the propertiesremain untouched. Therefore, we have an induced system of divided powers on H∗A.

6.3(b): Consider the divided polynomial algebra Γ(y) with degy = 2, and assume that y is a bound-ary, y = ∂x for some x. I claim that y(i) is not a boundary. Indeed, suppose y(i) = ∂f for someelement f =

∑j zjy

(j) ∈ Γ(y), where zj ∈ Z. Then yi/i! =∑j zj∂(yj)/j! =

∑j zj [jy

j−1∂y]/j! =

∂y∑j zjy

j−1/(j − 1)! = ∂2x · w = 0 · w = 0. But this implies yi = 0, a contradiction.

6.4(a): By Theorem V.6.4[1] we have an injection ψ :∧∗

(G ⊗ k) → H∗(G, k) for G abelian and ka PID. This k-algebra map was the unique extension of the isomorphism G ⊗ k → H1(G, k) in dimen-sion 1. Now ψ[(g ⊗ 1) ∧ (h ⊗ 1)] = ψ(g ⊗ 1) · ψ(h ⊗ 1) by definition of an algebra map, where · isthe Pontryagin product. On the bar resolution this product is given by the shuffle product, and theisomorphism ψ : G ⊗ k → H1(G, k) sends g ⊗ 1 to [g]. Thus (g ⊗ 1) ∧ (h ⊗ 1) 7→ [g] · [h] =

∑σ[g|h] =

[g|h]− [h|g], where σ ran over the two possible (1, 1)-shuffles. Remark: This map is well-defined because

for (g ⊗ 1) ∧ (g ⊗ 1) = 0 ∈∧2

(G⊗ k), the image is [g|g]− [g|g] = 0.

6.4(b): Let k be a PID in which 2 is invertible, let G be an abelian group, and consider the mapC2(G, k)→

∧∗(G⊗k) given by [g|h] 7→ (g⊗1)∧(h⊗1)/2. This induces a map ϕ : H2(G, k)→

∧∗(G⊗k)

because any 3-coboundary ∂[r|s|t] = [s|t] − [rs|t] + [r|st] − [r|s] is mapped to the trivial element(s⊗1)∧(t⊗1)/2−(r+s⊗1)∧(t⊗1)/2+(r⊗1)∧(s+ t⊗1)/2−(r⊗1)∧(s⊗1)/2 = [(s⊗1)∧(t⊗1)/2]−[(r⊗1)∧(t⊗1)/2]− [(s⊗1)∧(t⊗1)/2]+[(r⊗1)∧(s⊗1)/2]+[(r⊗1)∧(t⊗1)/2]− [(r⊗1)∧(s⊗1)/2] = 0.Using ψ from part(a) above, ϕ is its left-inverse because (g⊗ 1)∧ (h⊗ 1) 7→ [g|h]− [h|g] 7→ (g⊗ 1)∧ (h⊗1)/2− (h⊗ 1) ∧ (g ⊗ 1)/2 = (g ⊗ 1) ∧ (h⊗ 1)/2 + (g ⊗ 1) ∧ (h⊗ 1)/2 = (g ⊗ 1) ∧ (h⊗ 1), where we notein the last equality that the exterior algebra is strictly anti-commutative.

6.5: Let G be abelian and let A be a G-module with trivial G-action. In view of the isomorphismH2G ∼=

∧2G, the universal coefficient theorem gives us a split exact sequence 0 → Ext(G,A) →

H2(G,A)θ→ Hom(

∧2G,A) → 0. The isomorphism ψ :

∧2G → H2G is given by g ∧ h 7→ [g|h] − [h|g]

by Exercise V.6.4(a). Now the map β : H2(G,A)→ Hom(H2G,A) in the universal coefficient sequencesends the class ‖f‖ of the cocycle f to (β‖f‖)([g1|g2]) = f(g1, g2). Then θ is given by ‖f‖ 7→ β‖f‖ 7→(β‖f‖ ψ)(g ∧ h) = (β‖f‖)([g|h] − [h|g]) = f(g, h) − f(h, g), and this element is an alternating map.Thus θ coincides with the map θ in Exercise IV.3.8(c), and so we see that every alternating map comesfrom a 2-cocycle. It also follows that Ext(G,A) ∼= Eab(G,A), whence the name “Ext.”

46

6 Chapter VI: Cohomology Theoryof Finite Groups

2.1: Suppose |G : H| <∞ and thatM is a ZG-module with a relative injective resolutionQ, and η : M →Q0 is the canonical admissible injection (i.e. Q0 = CoindGHResGHM). If M is free as a ZH-module then Q0

is ZG-free by Corollary VI.2.2[1]. Since η is H-split, the exact sequence 0→Mη→ Q0 → Cokerη → 0 is

split-exact and hence Q0 ∼= M⊕Cokerη. Since M and Q0 are both ZH-free (a free ZG-module is ZH-freeby Exercise I.3.1), Cokerη is by definition stably free. Let us find this cokernel explicitly. The canonicalinjection is given by M → HomH(ZG,ResGHM) ∼= IndGHResGHM = ZG ⊗ZH ResGHM , m 7→ [s 7→ sm] 7→∑g∈G/H g⊗g−1m. But IndGHResGHM

∼= Z[G/H]⊗M with the mapping g⊗m 7→ g⊗gm, by Proposition

III.5.6[1]. Thus the canonical injection is given by m 7→∑g∈G/H g ⊗m. We can now consider M as a

free ZH-module, and without loss of generalization we can assume M = ZH (since direct sums commutewith tensor products). We can also regard Z[G/H] =

⊕g∈G/H Z[g] as a direct sum of integers via the

isomorphism Z[g] ∼= Z, g 7→ 1. Thus the H-split injection is η : ZH →⊕|G:H| Z ⊗ ZH ∼=

⊕|G:H| ZHwith the mapping m 7→ (g1, . . . , g|G:H|) ⊗m 7→ (1, . . . , 1) ⊗m 7→ (m, . . . ,m). The cokernel of this map

is⊕|G:H| ZH/ZH ∼=

⊕|G:H|−1 ZH, a free ZH-module. Indeed, For any finite direct sum⊕n

X, the

quotient of this group by its diagonal subgroup X = (x, . . . , x) is the direct sum⊕n−1

X becauseany element (x1, . . . , xn−1, xn)X is equivalent in the quotient group to (x1x

−1n , . . . , xn−1x

−1n , 1)X via the

element (x−1n , . . . , x−1

n , x−1n ). Since Cokerη is free as a ZH-module, we can apply Corollary VI.2.2[1] to

get an admissible injection Cokerη → Q1 with Q1 free as a ZG-module. Continuing in this way, weobtain the resolution Q as a complex of free ZG-modules.

3.1: Let G act freely on S2k−1, as on pg20[1]. From this we have a free resolution of Z over ZGwhich is periodic of period 2k:

· · · → C1 → C0 → C2k−1 → · · · → C1 → C0ε→ Z→ 0

where C∗ = C∗(X) and X ≈ S2k−1 is a free G-complex that makes each Ci finitely generated. Then byProposition VI.3.5[1] we can take the above resolution ε : C → Z, form the dual (backwards resolution)of its suspension ε∗ : Z∗ = Z → ΣC, and then splice together C and ΣC to form a complete resolutionF for G. This resolution is obviously periodic of period 2k because ε is periodic of period 2k and thesuspension just shifts the resolution (leaving the period unaltered) and the dual functor forms a periodicresolution of the same period.If G = 〈t〉 is finite cyclic of order n, and k = 1, then G acts by rotations on the circle (n vertices/edges)and we have a periodic resolution of period 2:

· · · → ZG t−1→ ZG N→ ZG t−1→ ZG ε→ Z→ 0where N = 1 + t + · · · + tn−1 is the norm element. Now ε(1) = 1, so by Proposition VI.3.4[1] the dual

ε∗ : Z→ ZG is given by ε∗(1) =∑g∈G g = N . The maps ZG t−1→ ZG and ZG N→ ZG are invariant under

the dual functor because HomG(ZG,ZG) ∼= ZG. Therefore, the explicit complete resolution is:

· · · → ZG t−1→ ZG N→ ZG t−1→ ZG N→ ZG t−1→ ZG N→ ZG t−1→ ZG→ · · · .

5.1: We have a natural map H∗ → H∗ which is an isomorphism in positive dimensions and anepimorphism in dimension 0. The cup product for both functors agrees in dimension 0 because `:H0(G,M) ⊗ H0(G,N) → H0(G,M ⊗ N) is the map MG ⊗ NG → (M ⊗ N)G induced by the in-

clusions MG → M and NG → N , and `: H0(G,M) ⊗ H0(G,N) → H0(G,M ⊗ N) is induced by

MG ⊗NG → (M ⊗N)G via the surjection H0 H0. From this compatibilism in dimension 0 we candeduce that the diagram

Hp(G,M)⊗Hq(G,M)`

//

Hp+q(G,M ⊗N)

Hp(G,M)⊗ Hq(G,M)`

// Hp+q(G,M ⊗N)

commutes for all p, q ∈ Z. Indeed, embed M in the (co)induced module M = Hom(ZG,M) ∼= Z ⊗M(noting that G is finite for our purposes) and let 0→M → M → C → 0 be the canonical Z-split exactsequence (see Exercise III.7.3). For any G-module N the sequence 0→M⊗N → M⊗N → C⊗N → 0 is

47

exact, and the module M ⊗N is induced (see Exercise III.5.2(b)). We therefore have dimension-shiftingisomorphisms δ : Hp(G,C)→ Hp+1(G,M) and δ : Hp(G,C ⊗N)→ Hp+1(G,M ⊗N) which commuteswith the cup product (see pg110[1]). The compatibilism for p = 0, q = 0 allows us to prove by ascendinginduction on p that the above diagram is commutative for p ≥ 0 and q = 0. Embedding N in an inducedmodule, we then see that it commutes for q ≥ 0. The scenario p, q < 0 is trivial because Hp = 0 forp < 0. Thus the cup product on H∗ is compatible with that defined originally on H∗, and the naturalmap H∗ → H∗ preserves products.

5.2(a): Let H∗(G)(p) be the p-primary component of H∗(G) = H∗(G,Z), so that we have H∗(G) =⊕p | |G| H

∗(G)(p). Since H∗(G)(p) is a subgroup of H∗(G), in order to show that it is an ideal in H∗(G)

it suffices to show that α ` β ∈ H∗(G)(p) for α ∈ H∗(G)(p) and β ∈ H∗(G). But this is trivial because

if psα = 0 then ps(α ` β) = (psα) ` β = 0 ` β = 0. Now⊕

q 6=p H∗(G)(q) is also an ideal because

a sum of ideals is an ideal. Consequently, H∗(G)(p) is a quotient ring of H∗(G) via the projection

H∗(G) H∗(G)(p).

Note: The p-primary ring H∗(G)(p) is not a subring of H∗(G) because the inclusion H∗(G)(p) →H∗(G) does not preserve identity elements. The identity of H∗(G)(p) is 1 ∈ Zpr = H0(G)(p) while the

identity of H∗(G) is 1 = (1, . . . , 1) ∈ Z|G| = H0(G), where we have used the factorization of Z|G| into thedirect sum of its p-primary components and |G| = prm with p - m. The inclusion will send the identity1 to (1, 0, 0, . . . , 0) 6= 1, where we take the first summand of Z|G| to be Zpr .

5.2(b): Consider the group isomorphism ϕ : H∗(G)∼=→∏p | |G| H

∗(G)(p), where each factor on the

right is a ring via part(a), and multiplication in the product is done componentwise [note: we switchfrom direct sum to direct product notation in order to emphasize the fact that we are dealing cat-egorically with rings]. The map is given by ϕ(α) = (. . . , αp, . . .), where α =

∑p αp in the decom-

position H∗(G) =⊕

p | |G| H∗(G)(p). In order to show that this map is a ring isomorphism it suf-

fices to show that (αβ)p = αpβp, where αβ = α ` β, for then ϕ(α ` β) = (. . . , (αβ)p, . . .) =(. . . , αpβp, . . .) = (. . . , αp, . . .)(. . . , βp, . . .) = ϕ(α) ` ϕ(β). Writing α =

∑p αp and β =

∑p′ βp′ we

have αβ =∑p,p′ αpβp′ =

∑p αpβp, because αpβp′ 6=p is annihilated by both p and p′ 6= p, and hence

annihilated by 1 = gcd(p, p′) [note that gcd(p, p′) = mp + np′ for some m,n ∈ Z by the EuclideanAlgorithm]. It is now obvious that (αβ)p = (

∑q αqβq)p = αpβp.

6.1:

6.2: Let R be a ring, let C be a chain complex of finitely generated projective R-modules, andlet C be the dual complex HomR(C,R) of finitely generated projective right R-modules. For anyz ∈ (C⊗RC)n any any chain complex C ′, there is a graded map ψz : HomR(C,C ′) → C⊗RC ′ ofdegree n, given by ψz(u) = (idC⊗u)(z). Let z ∈ (C⊗RC)0 correspond to idC under the isomor-phism ϕ : C⊗RC → HomR(C,C) from Exercise VI.6.1. This element is indeed a cycle, becauseϕ−1(∂z) = D0ϕ0(z) = D0(idC) = d idC − (−1)0idC d = d − d = 0 implies that ∂z = 0 since ϕis an isomorphism. Then z = (zp)p∈Z, where zp ∈ C−p ⊗R Cp = (Cp)

∗ ⊗R Cp corresponds to (−1)pidCpunder the canonical isomorphism φ : (Cp)

∗ ⊗R Cp ∼= HomR(Cp, Cp) of Proposition I.8.3[1] given byc⊗c 7→ 〈c, x〉c. Indeed, denoting zp = c⊗c, the canonical isomorphism gives 〈c, x〉c = (−1)pidCp and the

isomorphism ϕ0 = (ϕ−pp)p∈Z then becomes (−1)−p·p〈c, x〉c = (−1)p2

(−1)pidCp = (−1)p(p+1)idCp = idCp ,agreeing with our choice for z. Now ψz is induced by maps ψpq : HomR(C−p, C

′q) → Cp ⊗R C ′q. Since

u ∈ HomR(C−p, C′q) is of degree p + q and idCp is in dimension p, these maps are clearly given by

ψpq(u) = (−1)(q+p)p(idCp ⊗ u)(z−p); see the definition of a map between completed tensor products on

pg137[1]. Then Exercise I.8.7 states that ϕ−1pq = ψϕ−1(idCp ) = ψr, where ψr : HomR(C−p, C

′q)→ Cp⊗RC ′q

is defined by ψr(u) = (idCp ⊗R u)(r). Since r = ϕ−1(idCp) = (−1)pqφ−1(idCp) = (−1)pq(−1)pz−p, we

have ϕ−1pq (u) = ψr(u) = (idCp ⊗R u)((−1)pq(−1)pz−p) = (−1)(pq+p2)(idCp ⊗ u)(z−p) = ψpq(u), noting

that (−1)p = (−1)p2

. Therefore, ψz is the inverse of the isomorphism ϕ : C⊗RC ′ → HomR(C,C ′) ofExercise VI.6.1.

48

6.3: Let G be a finite group, let F be an acyclic chain complex of projective ZG-modules, and letε′ : F ′ → Z be a complete resolution. Part(b) of Proposition VI.6.1[1] states that if F is of finite typethen ε′ ⊗M induces a weak equivalence F ⊗G(F ′ ⊗M)→ F ⊗GM = F ⊗GM . The proof of the propo-sition considers the dual F = HomG(F,ZG) and utilizes the fact that it is projective. However, if wedid not impose the finiteness hypothesis on F then its dual would not necessarily by projective. Indeed,the dual of an infinite direct sum is an infinite direct product, and HomG(

⊕∞ ZG,ZG) ∼=∏∞ ZG is

not ZG-projective. If it were ZG-projective, then by Exercise I.8.2 it would also be Z-projective (i.e.free abelian). But any subgroup of a free abelian group is free abelian by Theorem I.7.3[5], and

∏∞ Zis a subgroup of

∏∞ ZG which is not free abelian, giving the desired contradiction. Thus the finitenesshypothesis is necessary for the given proof – this does not guarantee that the finiteness hypothesis isnecessary for the statement of the proposition.

6.4:

7.1(a): Let ϕ : Hi(G,M)∼=→ H−1−i(G,M) be the isomorphism established in the proof of Proposi-

tion VI.7.2[1], which on the chain level has the inverse ϕ−1 : HomG(F,ZG)⊗GM → HomG(F,M) given

by u ⊗ m 7→ [x 7→ u(x) · m], and let z = ϕ(1) ∈ H−1(G,Z). For an arbitrary G-module coefficienthomomorphism h : M → N we have a commutative diagram

HomG(F,ZG)⊗GM

α

ϕ−1

// HomG(F,M)

β

HomG(F,ZG)⊗G Nϕ−1

// HomG(F,N)

where α(u ⊗ m) = u ⊗ h(m) and β(f) = h f , because βϕ−1(u ⊗ m) = β[u(x) · m] = h(u(x) · m) =u(x) · h(m) = ϕ−1(u⊗ h(m)) = ϕ−1α(u⊗m). Thus ϕ−1 is natural and hence so is ϕ. Since ϕ is naturalthe following diagram with short exact rows is commutative (suppressing the end 0’s)

HomG(F,M ′) //

ϕ

HomG(F,M) //

ϕ

HomG(F,M ′′)

ϕ

HomG(F,ZG)⊗GM ′ // HomG(F,ZG)⊗GM // HomG(F,ZG)⊗GM ′′

and so ϕ is compatible with connecting homomorphisms in long exact sequences by Proposition I.0.4[1].

7.1(b): By definition of z, ϕ and a z agree on 1 ∈ H0(G,Z) since 1 a z = z. If now M and

u ∈ H0(G,M) are arbitrary, there is a coefficient homomorphism Z → M such that 1 7→ u un-

der the induced map α : H0(G,Z) → H0(G,M), noting that this cohomology map is induced fromH0(G,Z) = ZG →MG = H0(G,M). Since the cap product is natural with respect to coefficient homo-morphisms we have a commutative diagram

H0(G,Z)az//

α

H−1(G,Z)

β

H0(G,M)az// H−1(G,M)

which defines β(z) = β(1 a z) = α(1) a z = u a z. Thus by naturality of ϕ from part(a) we havean analogous commutative diagram as above (replacing a z with ϕ), and this yields ϕ(u) = ϕα(1) =βϕ(1) = β(z) = u a z.

7.1(c): The maps ϕ and a z agree in dimension 0 (referring to the domain) by part(b), and ϕ isδ-compatible by part(a). Thus we can use dimension-shifting [the δ boundary isomorphisms] to deducethat ϕ and a z agree in all dimensions, up to sign. Indeed, we have the commutative diagram

H0(G,M)ϕ//

δ∼=

H−1(G,M)

δ∼=

Hn(G,K)ϕ// H−1−n(G,K)

49

where the vertical maps are due to iterations of the dimension-shifting technique 5.4 on pg136[1]. Theseisomorphisms provide ambiguity in the sign, so ϕ(u) = ±u a z in any dimension and hence PropositionVI.7.2[1] has been reproved (the isomorphism is given by the cap product with the fundamental class z).

7.2: Let A be an abelian torsion group, and consider the injective resolution 0→ Z→ Q→ Q/Z→ 0 of

Z. Applying Hom(A,−) gives the cochain complex 0→ Hom(A,Z)→ Hom(A,Q)δ0→ Hom(A,Q/Z)

δ1→ 0,and we have Ext(A,Z) ≡ Ext1

Z(A,Z) = Kerδ1/Imδ0 = Hom(A,Q/Z)/Imδ0 = A′/Imδ0. But Q is torsion-free, so Hom(A,Q) = 0 and Imδ0 = 0. Thus Ext(A,Z) = A′.Equivalently, Theorem 17.1.10[2] provides us with a long exact sequence 0→ Hom(A,Z)→ Hom(A,Q)→Hom(A,Q/Z) → Ext(A,Z) → Ext(A,Q). But Hom(A,Q) = 0 as mentioned above, and Ext(A,Q) = 0

by Proposition 17.1.9[2] sinceQ is Z-injective. Thus we have a desired isomorphismA′ = Hom(A,Q/Z)∼=→

Ext(A,Z).

7.3: Let G be a finite group, let M be a G-module which is free as an abelian group, and let F bea projective resolution of Z over ZG. Note that M∗ = Hom(M,Z) by Proposition VI.3.4[1]. Considerthe split exact coefficient sequence 0 → Z → Q → Q/Z → 0. Since M is Z-free, applying the functorHom(M,−) yields the exact sequence 0 → M∗ → Hom(M,Q) → M ′ → 0 where M ′ = Hom(M,Q/Z);this sequence is Z-split exact because the original coefficient sequence is split exact (Hom commutes

with direct sums). I claim that H∗(G,Hom(M,Q)) = 0 and H∗(G,Hom(M,Q) ⊗M) = 0. Assuming

this for the moment, we then have dimension-shifting isomorphisms δ : Hj(G,M ′) → Hj+1(G,M∗) forall j. It is a fact that the tensor product of a G-module with a Z-split exact sequence is exact, so0→M∗ ⊗M → Hom(M,Q)⊗M →M ′ ⊗M → 0 is an exact sequence. Thus we also have dimension-

shifting isomorphisms δ : Hj(G,M ′ ⊗M) → Hj+1(G,M∗ ⊗M). Moreover, we have a commutativediagram

Hi−1(G,M ′)⊗ H−i(G,M)

δ⊗id∼=

`// H−1(G,M ′ ⊗M)

δ∼=

α // H−1(G,Q/Z)

δ∼=

Hi(G,M∗)⊗ H−i(G,M)`// H0(G,M∗ ⊗M)

β// H0(G,Z)

where the left-side square follows from compatibility with δ (see pg110[1]) and the right-side squarefollows from naturality of the long exact cohomology sequence (see pg72[1]); α and β are induced by theevaluation maps. Since the top row is a duality pairing by Corollary VI.7.3[1], so is the bottom row.It suffices to prove the claim. The analog of Proposition III.10.1[1] for Tate cohomology states that if

Hn(H,M) = 0 for some n with H ⊆ G, then Hn(G,M) is annihilated by |G : H|. Taking H = 1and M a rational vector space, the norm map N : MH → MH is an isomorphism (N = idM ). Then

H−1(1,M) = KerN = 0 = CokerN = H0(1,M), so H−1(G,M) and H0(G,M) are annihilated by|G| and are thus trivial groups since |G| is invertible in M . The claim is now justified since Hom(M,Q)and Hom(M,Q)⊗M are both rational vector spaces.

7.4: Let k be an arbitrary commutative ring and Q an injective k-module. Let A′ = Homk(A,Q) for any

k-module A. If M is a kG-module, then the pairing Hi(G,M ′)⊗ H−1−i(G,M)`→ H−1(G,M ′ ⊗M)→

H−1(G,Q) → Q induces an isomorphism Hi(G,M ′) ∼= H−1−i(G,M)′. Indeed, this is simply the analogof Corollary VI.7.3[1], and the proof of that corollary goes through untouched if we replace Q/Z by Q(as both are injective) and Z by k (as both are commutative ring coefficients).

8.1: Let G be a group and M a ZG-module such that H∗(G,M) = 0 but M is not cohomologicallytrivial. If G is cyclic, then by Theorem VI.8.7[1] it cannot be a p-group, so G = Zp×Zq ∼= Zpq for distinct

primes p and q. The complete resolution from Exercise VI.3.1 then implies that Hn(G,M) = CokerN

for n even and Hn(G,M) = KerN for n odd (see pg58[1]), so N : MG → MG is an isomorphism. Also,

Proposition VI.8.8[1] implies that either Hi(Zp,M) 6= 0 for some i or Hj(Zq,M) 6= 0 for some j (or

both). As mentioned on pg150[1] as a consequence of Theorem VI.8.5[1], Hi(Zp,Zp) 6= 0 for all i > 0.Therefore, let us consider M = Zp. It suffices to find a Zpq-action on Zp such that N : (Zp)Zpq → (Zp)Zpqis an isomorphism. But Zp is simple, so either Zpq acts trivially on Zp or (Zp)Zpq = (Zp)Zpq = 0. But if

50

Zpq acts trivially on Zp then N : Zp → Zp is the zero map (hence has nontrivial kernel) since multipli-cation by |Zpq| = pq annihilates Zp. Thus we must have (Zp)Zpq = (Zp)Zpq = 0. Taking p to be an oddprime, this condition is satisfied by the Zpq-action x ·mi = m2i, where Zp = 〈m〉 and Zpq = 〈x〉, as longas 2pq ≡ 1 mod p (because we must have m = xpq ·m = m2pq ). For example, p = 3, q = 2 works, asdoes p = 7, q = 3.As a result, a desired example is the group G = Z6 = 〈x〉 and the Z6-module M = Z3 = 〈m〉 coupledwith the action x ·mi = m2i.

8.2: Suppose M is a G-module which is Z-free and cohomologically trivial (G is of course finite).Then M is ZG-projective by Theorem VI.8.10[1], so F = M ⊕N for some projective module N and freemodule F. For any G-module K, the module Hom(F,K) is an induced module by Exercise III.5.2(b)(since F ∼= ZG⊗F′ = IndG1F

′ where F′ is a free Z-module of the same rank) and hence cohomologicallytrivial. Since the Hom-functor commutes with direct sums, Hom(M,K) is also cohomologically trivialfor any G-module K.Alternatively, for any G-module K choose an exact sequence 0 → L → F → K → 0 with F free (suchsequences exist because every module is a quotient of a free module). Since M is Z-free, we can applyHom(M,−) to get the exact sequence 0 → Hom(M,L) → Hom(M,F ) → Hom(M,K) → 0. Since Fand L are also Z-free, Hom(M,L) and Hom(M,F ) are cohomologically trivial by Lemma VI.8.11[1]; Lis free because it embeds in the free Z-module F and any submodule of a Z-free module is free. ThusHom(M,K) is cohomologically trivial by the long exact Tate cohomology sequence.

8.3(a): Let M and P be ZG-modules such that M is Z-free and P is ZP -projective, and consider

any exact sequence 0 → Pi→ E

ϕ→ M → 0. The obstruction to splitting the sequence lies inH1(G,Hom(M,P )) ∼= Ext1

ZG(M,P ), where the isomorphism follows from Proposition III.2.2[1]. Moreprecisely, we have a short exact sequence ofG-modules 0→ Hom(M,P )→ Hom(M,E)→ Hom(M,M)→0 since M is Z-free, and this yields the sequence HomG(M,E) → HomG(M,M)

δ→ H1(G,Hom(M,P ))via the long exact cohomology sequence, where we recall that HomG(·, ·) = Hom(·, ·)G = H0(G,Hom(·, ·)).Hence the extension splits iff δ(idM ) = 0, because if the extension splits then there is a section s : M → Ewhich maps onto idM (i.e. s 7→ ϕ s = idM ) and so idM ∈ Kerδ by exactness, and if δ(idM ) = 0 thenby exactness there exists a map M → E which maps onto idM and that map is then the desired section.It suffices to show that Hom(M,P ) is cohomologically trivial, for then H1(G,Hom(M,P )) = 0 andδ = 0. By additivity [Hom commutes with direct sums], it suffices to show that Hom(M,F) is cohomo-logically trivial for any free ZG-module F, since the projective P is a direct summand of some F. ButF ∼= CoindG1F

′ where F′ is a free Z-module of the same rank (by Corollary VI.2.3[1]), so Hom(M,F) isinduced (by Exercise III.5.2(b)) and hence cohomologically trivial.Alternatively, since M is Z-free the original exact sequence in consideration is Z-split (see ExerciseAE.27), so the injection i : P → E of G-modules is a Z-split injection, hence admissible. Since P isG-projective, it is relatively injective by Corollary VI.2.3[1] and so the mapping problem

Pi //

idP

E

f

Pcan be solved (i.e. there exists a map f : E → P such that f i = idP ). But this just means that

f : E → P is a ZG-splitting homomorphism for the sequence 0 → Pi→ E

ϕ→ M → 0, and so thissequence splits.

8.3(b): Let M be a ZG-module such that proj dimM < ∞, and consider the projective resolution0→ Pn → · · · → P0 → M → 0. We can break this up into short exact sequences Zi → Pi → Zi−1 → 0,where Zi is the kernel of Pi → Pi−1. Now Zi (i ≥ 0) is Z-free because it is a submodule of aZG-projective module which is a submodule of a ZG-free module F, and F is also necessarily Z-free and any subgroup (in particular, Zi) of a Z-free group is Z-free. Therefore, for the sequence0→ Pn = Zn−1 → Pn−1 → Zn−2 → 0 with Pn G-projective and Zn−2 Z-free, part(a) above implies thatthis sequence splits and hence Pn−1

∼= Pn⊕Zn−2. Since Pn−1 is G-projective, so is Zn−2. One now seesby descending induction on i that Zi is G-projective for i ≥ 0, so that 0 → Z0 → P0 → Z−1 = M → 0

51

is a projective resolution of length 1. Thus proj dimM ≤ 1.

8.4: Let G be a group such that there exists a free, finite G-CW-complex X with H∗(X) ∼= H∗(S2k−1).

Since Hi(S2k−1) = 0 for i 6= 0 and i 6= 2k−1, the augmented cellular chain complex C∗ = C∗(X) is a free

resolution of Z over ZG up to dimension 2k− 1. From the chain sequence C2k−1∂2k−1→ C2k−2

∂2k−2→ C2k−3

we have Ker∂2k−2 = Im∂2k−1∼= C2k−1/Ker∂2k−1 [the isomorphism is due to the 1st Isomorphism

Theorem] and hence we have an exact sequence 0 → C2k−1/Ker∂2k−1 → C2k−2∂2k−2→ C2k−3 → · · · .

Now Im∂2k = B is the module of (2k − 1)-boundaries of C∗, and we have a surjection C2k−1/B C2k−1/Ker∂2k−1 with kernel Ker∂2k−1/B ∼= Z [note: this isomorphism comes from the fact that Z ∼=H2k−1(S2k−1)]. Thus we have an exact sequence S of G-modules 0 → Z → C2k−1/B → C2k−2

∂2k−2→C2k−3 → · · · → C0

ε→ Z→ 0 where each Ci is G-free. I claim that C2k−1/B is Z-free and has finite projec-tive dimension. Assuming this claim holds, C2k−1/B is cohomologically trivial by Theorem VI.8.12[1] andhence is ZG-projective by Theorem VI.8.10[1]. Thus we can splice together an infinite number of copies ofS (which forms an acyclic complex of projective G-modules) and we can then apply Proposition VI.3.5[1]to obtain a complete resolution which is periodic of period 2k. It suffices to prove the claim. Since C2k−1

is ZG-free, it is necessarily Z-free and hence any subgroup (in particular, B) must also be Z-free; thusC2k−1/B is Z-free. As X is a finite complex, C∗(X) stops after Cn(X) for some 2k − 1 < n < ∞.

Thus we have a finite projective resolution 0 → Cn∂n→ · · · → C2k

∂2k→ C2k−1 → C2k−1/B → 0 andproj dimC2k−1/B <∞.

9.1: Let G be a nontrivial finite group which has periodic cohomology of period d. Then there isan element u ∈ Hd(G) which is invertible in the ring H∗(G), so cup product with u gives a peri-odicity isomorphism v 7→ u ` v. Taking v = u and using anti-commutativity of the cup product,u ` u = (−1)d

2

(u ` u). If d is not even, then 2u2 = 0. If |G| = 2 then G is cyclic (of order 2) and hencehas period d = 2 (which is even), so we must have |G| ≥ 3. But then 2u2 = 0 implies that u ` u = u2 = 0and hence u = 0 by the periodicity isomorphism. This contradicts the fact that u is nontrivial (it isinvertible), so d must be even.

9.2: If G is cyclic then H∗(G) is periodic of period 2 because G admits a 2-dimensional fixed-point-

free representation as a group of rotations (see pg154[1]); we could also just note that Hi(G) is Z|G|for i even and is 0 for i odd. Conversely, if a group G has periodic cohomology of period 2, thenGab = H1G = H−2(G) ∼= H0(G) = Z|G|. Now |G| = |Gab| = |G|/|[G,G]| ⇒ |[G,G]| = 1 and henceG = Gab ∼= Z|G|, which is cyclic.

9.3: Suppose H∗(G) is periodic of period 4. We have H−1(G) = 0 and H0(G) = Z|G|, as explained on

pg135[1]. We also have H−2(G) = H1G = Gab, and H1(G) = H1G = Hom(G,Z) = 0 by Exercise III.1.2

(noting that G is finite by hypothesis). Thus, since Hn(G) ∼= Hn+4(G) for all n,

Hn(G) ∼=

Z n = 0

Gab n ≡ 1 mod 4

Z|G| n ≡ 3 mod 4

0 otherwise

Hn(G) ∼=

Z n = 0

Z|G| n ≡ 0 mod 4 , n 6= 0

Gab n ≡ 2 mod 4

0 otherwise

Note that this reproves Exercise II.5.7(a), because the finite subgroup G ⊂ S3 ⊂ H∗ has H∗(G) peri-odic of period 4 (see pg155[1]) and hence H2G = 0.

9.4: Suppose the finite group G has periodic cohomology. Now H∗(G) ∼=∏p | |G| H

∗(G)(p) by Exer-

cise VI.5.2(b), and each factor H∗(G)(p) embeds in H∗(Sylp(G)) by the Tate cohomology version of

Theorem III.10.3[1], so Hi(G) = 0 for i odd if Hi(Sylp(G)) = 0 for each prime p. Thus to show that

Hi(G) = 0 for i odd it suffices to do this when G is a p-group. For the p-group G (of order pr) withperiodic cohomology, G is either a cyclic group Zpr or a generalized quaternion group Q2r by Proposition

VI.9.3[1]. If G is cyclic then it has period 2 by Exercise VI.9.2, so since H−1(G) = 0, Hi(G) = 0 for iodd. If G is generalized quaternion then it has period 4 as explained on pg155[1] (it is a finite subgroup

of H∗), so by Exercise VI.9.3, Hi(G) = 0 for i odd.

52

9.5: Suppose G is a p-group which has a unique subgroup C of order p; note that C is necessarilycyclic. Choose a fixed-point-free representation of C on a 2-dimensional vector space W (as a group ofrotations), and form the induced module V = ZG ⊗ZC W . Since C is unique, it is normal in G andhence ResGCV

∼=⊕

g∈G/C gW by Proposition III.5.6[1]; each gW is clearly a fixed-point-free representa-

tion of C (c · gw = (cg)w 6= gw). Consequently, V is a fixed-point-free represenation of C. But thenV is also a fixed-point-free representation of G, for a nontrivial isotropy group Gv(v ∈ V − 0) wouldcontain an element x of order p (by Cauchy’s Theorem, Theorem 3.2.11[2]) and hence would containC (uniqueness implies C = 〈x〉), contradicting the fact that C acts freely on V − 0. Thus G admitsa periodic complete resolution of period 2|G : C| as explained on pg154[1], so G has periodic cohomology.

9.6: Let G = Zm o Zn, where m and n are relatively prime and Zn acts on Zm via a homomor-phism Zn → Z∗m whose image has order k. If a prime q divides n, then a Sylow q-subgroup H liesin Zn and is necessarily central (in Zn) because cyclic groups are abelian. By Theorem III.10.3[1] we

have H∗(Zn)(q)∼= H∗(H)Zn/H and by Exercise III.8.1 we know that Zn/H acts trivially on H∗(H),

so H∗(H) ∼= H∗(Zn)(q). By Theorem III.10.3[1] we also have H∗(G)(q) ⊆ H∗(H) ∼= H∗(Zn)(q). Since

H∗(Zn)(q) ⊆ H∗(G)(q) by Exercise AE.55, we must have H∗(G)(q)∼= H∗(Zn)(q). Now if a prime p divides

m, then the same argument yields H∗(H) ∼= H∗(Zm)(p) where H is now a Sylow p-subgroup of Zm ⊂ G.We have H/G because H is the unique subgroup of Zm/G (hence gHg−1 ∼= H for all g ∈ G), so Theorem

III.10.3[1] implies H∗(G)(p)∼= H∗(H)G/H ∼= H∗(Zm)

G/H(p) = H∗(Zm)

(Zm/H)oZn(p) . We have a Zn-action

on Zm given by ϕ : Zn → Aut(Zm) = Z∗m, and we have a trivial Zm/H-action ψ : Zm/H → Aut(Zm)given by ψ(g) = idZm ≡ 1. We then have a (Zm/H) o Zn-action ψ o ϕ which is precisely the actionϕ : Zn → Z∗m because (ψoϕ)(g, z) = ψ(g) ·ϕ(z) = 1 ·ϕ(z) = ϕ(z). Thus we can consider the G/H-action

on Zm (hence on its Tate cohomology) as the Zn-action, so H∗(G)(p)∼= H∗(Zm)Zn(p). Since H∗(G) is the

direct sum of its primary components and p (resp. q) ranges over prime divisors of m (resp. n), we have

the isomorphism H∗(G) ∼= H∗(Zn)⊕ H∗(Zm)Zn .Let us examine the Zn-action a little more carefully. The image of Zn under the action-homomorphismconsists solely of automorphisms f : Zm → Zm such that fk = idZm . Such a map induces f∗ = H2(f) on

H2(Zm) ∼= Zm. If α generates the 2nd-dimension cohomology, then in the cohomology ring, f∗(α) = λα

for some λ ∈ Zm. But α = id∗(α) = fk∗ (α) = λkα, so λk ≡ 1 modm. Noting that Hi(Zm) = 0 for iodd, but is nontrivial for i even. Now f∗(α

i) = f∗(α ` · · · ` α) = f∗(α) ` · · · ` f∗(α) = (λα) ` · · · `(λα) = λiαi which is the identity only when i is a multiple of 2k. Thus Hi(Zm)Zn is nontrivial onlywhen i ∈ 2kZ (in which case it is Zm since the action is trivial). Therefore,

Hi(Zm o Zn) ∼= Hi(Zn)⊕ Hi(Zm)Zn ∼=

Zmn i ≡ 0 mod 2k

0 i odd

Zn otherwise

for the case gcd(m,n) = 1. This means that the period of H∗(G) is 2k.

9.7: Let Fq be a field with q elements, where q is a prime power. The special linear group G = SLn(Fq) isthe kernel of the surjective determinant homomorphism det : GLn(Fq)→ F∗q , i.e. it is the group of matri-ces with determinant 1. Let us first assume that n ≥ 3. Then the cyclic groupsA = 〈diag(a, a−1, 1, . . . , 1)〉and B = 〈diag(1, . . . , 1, b, b−1)〉 form a non-cyclic abelian subgroup A×B ⊆ G, noting that Fq is commu-tative. If we now let n = 2 then we will assume that q is not prime. Then the cyclic groups A = 〈

(1 a0 1

)〉

and B = 〈(

1 b0 1

)〉 form a non-cyclic abelian subgroup A×B ⊆ G, since

(1 a0 1

)(1 b0 1

)=(

1 a+b0 1

)with b not

equal to any multiple of a (and vice versa). By Theorem VI.9.5[1], G = SLn(Fq) does not have periodiccohomology if n ≥ 3 or if q is not prime.Note that if n = 2 and q is prime then SL2(Fq) does have periodic cohomology, as explained on pg157[1].

9.8:

9.9: Suppose that G has p-periodic cohomology. Let P ⊆ G be a subgroup of order p, let N(P )(resp. C(P )) be the normalizer (resp. centralizer) of P in G, and let W = N(P )/C(P ); note that

53

if C(P ) = P then W is called the Weyl group. Choose a Sylow p-subgroup H containing P . SinceG has p-periodic cohomology, H is either cyclic or generalized quaternion by Theorem VI.9.7[1]. Iassert that H ⊆ C(P ) ⊆ N(P ). Indeed, if H is cyclic then it is necessarily abelian so the result fol-lows, and if H is generalized quaternion (must have p = 2) then it has a unique element of order 2(as stated on pg98[1]) and this is then the generator for P ∼= Z2 which must be central in H, so the

result follows. Denote by XG the G-invariant elements of H∗(H,M) and similarly for XN(P ), where

an element z ∈ H∗(H,M) is G-invariant if resHH∩gHg−1z = resgHg−1

H∩gHg−1gz for all g ∈ G. Theorem

III.10.3[1] states that H∗(G,M)(p)∼= XG and H∗(N(P ),M)(p)

∼= XN(P ). Now trivially, XG ⊆ XN(P ),

so it suffices to show that XN(P ) ⊆ XG, for then H∗(G,M)(p)∼= XG = XN(P )

∼= H∗(N(P ),M)(p).Note that P is the unique subgroup in H of order p because if H is generalized quaternion then thereasoning is as stated above and if H is cyclic then every subgroup has unique order (by Theorem2.3.7[2]). If H ∩ gHg−1 is trivial then every element in H∗(H,M) is clearly invariant for such g ∈ G.If H ∩ gHg−1 is not trivial then its order is at least p and the intersection contains P . This impliesP ⊆ gHg−1 ⇒ g−1Pg ⊆ H ⇒ g−1Pg = P and hence g ∈ N(P ), so the question of G-invariancereduces to the question of N(P )-invariance, i.e. XN(P ) ⊆ XG.Also, XN(P ) ⊆ XC(P ) trivially, so by Theorem III.10.3[1] we have the inclusion (up to isomorphism)

H∗(N(P ),M)(p) ⊆ H∗(C(P ),M)(p). Since H is a Sylow p-subgroup contained in C(P ), |W | and

p are relatively prime and hence corN(P )C(P ) res

N(P )C(P ) = |N(P ) : C(P )| = |W | is an isomorphism on

H∗(N(P ),M)(p), where the equality is due to Proposition III.9.5[1]. Thus the restriction map induces

a monomorphism H∗(N(P ),M)(p) → H∗(C(P ),M). But as explained on pg84[1], if z = resN(P )C(P )u

then z is N(P )-invariant; let the N(P )-invariants be denoted by Y ⊆ H∗(C(P ),M). Thus resN(P )C(P )

maps H∗(N(P ),M)(p) monomorphically into Y . Since C(P ) / N(P ), Y = H∗(C(P ),M)W as noted

on pg84[1]. Thus H∗(N(P ),M)(p) ⊆ H∗(C(P ),M)W , and so combining the two inclusions we see that

H∗(N(P ),M)(p) ⊆ H∗(C(P ),M)(p) ∩ H∗(C(P ),M)W = H∗(C(P ),M)W(p). For the other direction, if

z ∈ Y(p) = H∗(C(P ),M)W(p) then consider the element w = corN(P )C(P )z. Since H∗(C(P ),M)(p) is annihi-

lated by a power of p, w ∈ H∗(N(P ),M)(p). Regurgitating the proof of Theorem III.10.3[1] using the

double-coset formula, we deduce that z = resN(P )C(P )w

′ where w′ = w/|W | ∈ H∗(N(P ),M)(p). This means

that H(C(P ),M)W(p) ⊆ H∗(N(P ),M)(p) because resN(P )C(P ) maps H∗(N(P ),M)(p) monomorphically into

the N(P )-invariants. Thus H∗(G,M)(p)∼= H∗(N(P ),M)(p)

∼= H∗(C(P ),M)W(p).

9.10: For any finite group G, the augmentation ideal I ⊂ ZG is a cyclic G-module if G is cyclicgroup by Exercise I.2.1(b); we could also just note that if G = 〈s〉 then I = ZG · (s − 1) because Iconsists of elements of the form sk − 1 [we then form the elements si − sj ∈ I via summation], andsk − 1 = N · (s− 1) where N = sk−1 + · · ·+ s+ 1 ∈ ZG. Conversely, if I is cyclic as a G-module, so thatI = ZG · x, then I claim that G admits a periodic resolution of period 2. Assuming this for the moment,G then has periodic cohomology (of period 2) by Theorem VI.9.1[1] and hence G is cyclic by ExerciseVI.9.2. It suffices to prove the claim. The multiplication map ZG → ZG given by r 7→ rx has image Iand kernel K, so we can form the exact sequence 0→ K → ZG x→ ZG ε→ Z→ 0. Under the category ofabelian groups, the Rank-Nullity Theorem gives |G| = dimZZG = dimZI + dimZZ = dimZI + 1 for theaugmentation map ε, and gives |G| = dimZZG = dimZK+dimZI for the x-multiplication map. The firstequation implies dimZI = |G| − 1 and the second equation then implies dimZK = 1, i.e. K ∼= Z as anabelian group. So with K = 〈k〉, G acts on K via gk0 = zk0 (for z ∈ Z). But then k0 = g|G|k0 = z|G|k0

and hence z = 1, i.e. the G-action is trivial. Our exact sequence is now 0→ Z→ ZG x→ ZG ε→ Z→ 0.Splicing together this sequence infinitely many times, we obtain the desired periodic resolution.

54

7 Chapter VII: Equivariant Homology andSpectral Sequences

2.1: Let 0 → C ′ → C → C ′′ → 0 be a short exact sequence of chain complexes, and let FpC bethe filtration such that F0C = 0, F1C = C ′, and F2C = C. Then E1

1q = H1+q(C′/0) = H1+q(C

′) andE1

2q = H2+q(C/C′) = H2+q(C

′′) and E1pq = 0 for p ≤ 0 and p > 2. The differential d1 : E1

pq → E1p−1,q

gives maps ∂ : Hn(C ′′)→ Hn−1(C ′) with n = 2 + q and ∂ = d12q.

Now E1 is given by · · · → E13q = 0

d13→ E12q

∂→ E11q

d11→ E10q = 0→ · · · , and from E2 = H(E1) we see that

E21q = Kerd1

1/Im∂ = H(C ′)/Im∂ = Coker∂ and E22q = Ker∂/Imd1

3 = Ker∂ and E2pq = 0 for p ≤ 0 and

p > 2. In E2 we have Ker∂d22q→ 0 and Coker∂

d21q→ 0 and trivial maps for p 6= 1 and p 6= 2, because thedifferential d2 : E2

pq → E2p−2,q+1 is of bidegree (−2, 1). Thus H(E2) = E2, and so the spectral sequence

“collapses” at E2 (i.e. E2 ∼= E∞) because Er+1 ∼= H(Er) in general.

Since E∞p = GrpH(C) := FpH(C)/Fp−1H(C) where FpH(C) = ImH(FpC)ϕ→ H(C), we see that

E∞p = 0 for p ≤ 0 and p > 2 and Coker∂ = E2 ∼= E∞1 = F1H(C) and Ker∂ = E2 ∼= E∞2 =F2H(C)/F1H(C). Since F2H(C) = H(C) we have a surjection Hn(C) F2H(C)/F1H(C) ∼= Ker∂,

and since F1H(C) = ImH(C ′)ϕ→ H(C) we have a commutative diagram

Hn(C ′) //

ϕ%% %%

Hn(C)

F1Hn(C)?

OO

hence an injection Coker∂ = F1Hn(C) → Hn(C). We can now rewrite the E∞ sequence 0→ F1Hn(C)→F2Hn(C)/F1Hn(C) → 0 as a short exact sequence 0 → Coker∂ → Hn(C) Ker∂ → 0, and we thenhave a commutative diagram

Hn(C ′)i //

ϕ%% %%

Hn(C)j

//

$$ $$

Hn(C ′′)

Coker∂+

99

Ker∂,

::

The top row is exact at Hn(C) because Kerj = KerHn(C) Ker∂ = Coker∂ = Imϕ = Imi. SinceImj = Ker∂ and Keri = Kerϕ = Im∂, the top row of this commutative diagram extends to a long exacthomology sequence.We have thus deduced the familiar long exact homology sequence from this spectral sequence, and thespectral sequence of a filtered complex can be regarded as a generalization of the long exact sequenceassociated to a chain complex and a subcomplex.

3.1(a): Let C be a first-quadrant double complex such that the associated spectral sequence to Fp(TC)n =⊕i≤p Ci,n−i has E1

pq = 0 for q 6= 0, and let D be the chain complex E1∗,0 with differential d1. We

have the isomorphisms Hn(TC) = Ker[⊕

p≤n Cp,n−p →⊕

p≤n−1 Cp,n−1−p]/Im[⊕

p≤n+1 Cp,n+1−p →⊕p≤n Cp,n−p] = [

⊕p≤n−1Hn−p(Cp,∗)]⊕X = [

⊕p≤n−1E

1p,n−p]⊕X = 0⊕X, whereX = Cn,0/(Im[Cn+1,0 →

Cn,0] ⊕ Im[Cn,1 → Cn,0]) = Hn(C∗,0/Im[C∗,1 → C∗,0]) = Hn(H0(C∗,∗)) = Hn(E1∗,0) = Hn(D). Thus

Hn(TC) ∼= Hn(D).

3.1(b): Take τ : TC → D to be the canonical surjection, and note that this can be viewed as amap of double complexes C → D (where D is regarded as a double complex concentrated on the lineq = 0); this is obviously a filtration-preserving chain map. Now E1

pq(D) = Hq(E1p,∗ ≡ E1

p,0) which is 0 ifq 6= 0, and is Ker∂0 ≡ E1

p,0 if q = 0 (since E1p,−1 = 0). This is precisely the spectral sequence associated to

C, so the induced map on spectral sequences from τ is an isomorphism at the E1-level. Thus by Propo-sition VIII.2.6, τ induces an isomorphism H∗(TC)→ H∗(TD) ∼= H∗(D) and hence is a weak equivalence.

4.1: Suppose X is the union of subcomplexes Xα such that every non-empty intersection Xα0∩· · ·∩Xαp

(p ≥ 0) is acyclic, and let K be the nerve of the covering Xα. Let C be the double complexCpq =

⊕σ∈K(p) Cq(Xσ), and let T be the [total] chain complex TC. As shown on pg167[1], we have

a spectral sequence with E1pq equal to 0 if q 6= 0 and equal to Cp(X) if q = 0. Then Exercise 3.1(b)

55

implies that we have a weak equivalence T → C(X). Moreover, we have another spectral sequence withE1pq = Cp(K,Hq) where Hq ≡ Hq(Xσ) is a coefficient system on K. Since each Xσ is acyclic, E1

pq is0 for q 6= 0 and is Cp(X,Z) for q = 0. Then Exercise 3.1(b) implies that we have a weak equivalenceT → C(K). Thus we have an isomorphism H∗(X) ∼= H∗(K).

7.2: Let X be a G-complex such that for each cell σ of X, the isotropy group Gσ fixes σ pointwise; in thiscase the orbit space X/G inherits a CW-structure. Assume further that each Gσ is finite. We have a spec-tral sequence E1

pq =⊕

σ∈ΣpHq(Gσ,Mσ)⇒ HG

p+q(X,M), where Σp is a set of representatives for Xp/G

(Xp is the set of p-cells of X) and Mσ = Zσ ⊗M (Zσ is the Gσ-module additively isomorphic to Z, onwhich Gσ acts by the orientation character χσ : Gσ → ±1). Since Gσ fixes σ pointwise, χσ(Gσ) = 1and hence Zσ = Z and hence Qσ = Z ⊗ Q = Q (where we now take rational coefficients M = Q). Nowfor all q > 0, Hq(Gσ,Q) = Hq(Gσ)⊗Q = 0 where the first equality is proved in Exercise AE.6 and thelatter equality follows from the fact that Hq(Gσ) is finite (proved in Exercise AE.16). The E1 term isthen concentrated on the line q = 0, and the spectral sequence therefore collapses at E2 = H∗(E

1) to

yield HG∗ (X,Q) ∼= H∗(

⊕σ∈Σp

H0(Gσ,Qσ))∗= H∗(H0(G,

⊕σ∈Σp

IndGGσQσ)) = H∗(H0(G,Cp(X;Q))) =

H∗(Cp(X;Q)G) = H∗(X/G;Q), where the starred equality is the result of Shapiro’s lemma and the lastequality is given by Proposition II.2.4[1].If X is also contractible, then X is necessarily acyclic. Thus the above result and Proposition VII.7.3[1]imply H∗(G,Q) ∼= HG

∗ (X,Q) ∼= H∗(X/G;Q).Note: The hypothesis that Gσ fixes σ pointwise is not very restrictive in practice. In the case of a

simplicial action, it can always be achieved by passage to the barycentric subdivision X. Indeed, forσ′ ⊂ X which lies in σ ⊂ X, Gσ′ ⊆ Gσ. If Gσ′ did not fix σ′ pointwise then this would break continuityof the G-action on σ (consider two such simplices of X which lie in σ and have a common face).

7.3: Let X be a G-complex and E a free contractible G-complex. There is a G-map X × E → Xwhich is a homotopy equivalence, so Proposition VII.7.3 implies that HG

∗ (X × E) → HG∗ (X) is an iso-

morphism. As G acts freely on X × E, HG∗ (X × E) ∼= HG

∗ ((X × E)/G). Thus we have the equivalenceHG∗ (X) ∼= H∗(X ×G E).

7.5: Let X be a G-complex and let N be a normal subgroup of G which acts freely on X. Let Zσbe the orientation module and let Xp denote the set of p-cells of X. Let Σp be a set of representativesfor Xp/G and let Σ′p be a set of representatives for (X/N)p/(G/N); it is easy to see that both sets

are in bijective correspondence. To prove that HG∗ (X,M) ∼= H

G/N∗ (X/N,M) with any G/N -module

coefficient M , it suffices to show that⊕

σ∈ΣpHq(Gσ,Mσ) ∼=

⊕σ′∈Σ′p

Hq((G/N)σ′ ,Mσ′); this is because

we have a spectral sequence E1pq =

⊕σ∈Σp

Hq(Gσ,Mσ) ⇒ HGp+q(X,M) and so the said isomorphism

will give isomorphic associated graded modules (since Er = H(Er−1) and E∞ is the associated graded

module), and this will give HG∗ (X,M) ∼= H

G/N∗ (X/N,M) by Lemma VII.2.1[1]. In view of the bi-

jection Σp → Σ′p, σ 7→ σ′, it suffices to show that Hq(Gσ,Mσ) ∼= Hq((G/N)σ′ ,Mσ′). First notethat Mσ ≡ Zσ ⊗ M = Zσ′ ⊗ M ≡ Mσ′ because the Gσ-action and the (G/N)σ′ -action on Z coin-cide (if g ∈ Gσ preserves the orientation of σ ∈ X, gσ = +σ, then gN preserves the orientation ofσ′ = σ = nσ ∈ X/N because +σ′ = +σ = gσ = g · nσ = (gn)σ = (gn)σ′ ∈ X/N for all n ∈ N)and because the two actions coincide on M by definition (since M is a G/N -module). Thus it sufficesto show (due to the Kunneth formula) that Hq(Gσ) ∼= Hq((G/N)σ′) where the homology is now us-ing integer coefficients, and in turn it suffices to show that Gσ ∼= (G/N)σ′ where σ′ is the image ofσ under the quotient X → X/N . Consider the obvious monomorphism ϕ : Gσ → (G/N)σ′ given byg 7→ gN ; it suffices to show that ϕ is surjective. But this is immediate, because if gN ∈ (G/N)σ′

then gNσ′ = σ′ which implies gn1σ = n2σ for some n1, n2 ∈ N which implies n−12 gn1 ∈ Gσ, and then

ϕ(n−12 gn1) = n−1

2 gn1N = n−12 gN = n−1

2 g(g−1n2g)N = gN where we note that N / G.

10.1: The proof of Theorem VII.10.5[1] used the assumption that |G| = p (p prime) to state that

dimZpHn(G,Z)p = 1 for all n ∈ Z, because G ∼= Zp has Tate cohomology group Zp in every dimen-

sion. The proof also used that assumption in order to apply Proposition VII.10.1[1] to the G-invariantsubcomplex XG of X; the isotropy group Gσ for every cell σ ∈ X − Y cannot equal G (because XG isthe largest subcomplex on which G acts trivially) and hence must be the trivial group (the only proper

56

subgroup of G ∼= Zp).

10.2: The extended theorem holds for |G| = p1 by the original Theorem VII.10.5[1], so we proceedby induction, assuming the extended theorem holds for |G| = pr. Let |G| = pr+1 and remember thehypothesis that XH is a subcomplex for all H ⊆ G. Since G is a p-group we can choose a maximalnormal subgroup N of index p, so that XG = (XN )G/N . Via induction (since |N | = pr) we can apply theextended theorem to X with XN [which is subcomplex by hypothesis], so every condition of the theoremon X is also satisfied on XN . Since |G/N | = p we can apply the original theorem to XN with (XN )G/N

[which is a subcomplex since it is equal to XG], so every condition of the theorem on XN is also satisfiedon (XN )G/N . Thus every condition of the extended theorem on X is also satisfied on (XN )G/N = XG,and the proof is complete.

10.3: Let X be a finite-dimensional free G-complex (G finite) with H∗(X) ∼= H∗(S2k). Proposition

VII.10.1[1] (with Y = ∅) implies that HG∗ (X,M) = 0, noting that Gσ is trivial for all σ since X is

G-free. On the other hand, we have a spectral sequence E2pq = Hp(G,Hq(X,M)) ⇒ HG

p+q(X,M).Since the spectral sequence is concentrated on the horizontal lines q = 0 and q = 2k (the only

nonzero homology terms of the 2k-sphere), it follows that the differential d2k+1 : Hp(G,H2k(X,M)) →Hp−(2k+1)(G,H2k+(2k+1)−1(X,M)) = Hp−2k−1(G,H4k(X,M)) = 0 is an isomorphism. To prove thatevery nontrivial element of G acts nontrivially on H2k(X) it suffices to show that every cyclic sub-group of G (generated by the elements of G) acts nontrivially on H2k(X), so we are immediately re-duced to the case where G is cyclic and nontrivial. Using the isomorphism d2k+1 with M = Z and podd, we conclude that Hodd(G,H2k(X)) = 0. This means G acts nontrivially on H2k(X), otherwise

Hodd(G,H2k(X)) = Hodd(G,Z) = Z|G| which is not equal to 0. The proof is now complete. Note thata nontrivial G-action on H2k(X) ∼= Z means that G = Z2, so having every nontrivial element of G actnontrivially on H2k(X) means that |G| ≤ 2 (the case |G| = 1 is satisfied vacuously since there are nonontrivial elements).

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8 Chapter VIII: Finiteness Conditions

2.1: By definition, cd Γ = 0 iff Z admits a projective resolution 0→ P → Z→ 0 of length 0, i.e. Z ∼= Pand Z is ZΓ-projective. But Exercise I.8.1 states that only the trivial group Γ = 1 makes Z a projectivemodule. Thus the trivial group is the only group of cohomological dimension zero.

2.2: Take Γ = Z2. Then cd Γ = ∞ by Corollary 2.5. Now a free ZΓ-module F is a direct sum⊕ZΓ

where the Γ-action would be a parity-permutation on all or some of the summands, so Hn(Γ, F ) ∼=[⊕Hn(Γ, (ZΓ)2)] ⊕ [

⊕Hn(Γ,ZΓ)] where the first collection of summands has the nontrivial Γ-action.

But then that module (ZΓ)2 is an induced module IndΓ1ZΓ, and so are the other modules ZΓ (triv-

ially). So these modules are H∗-acyclic and hence Hn(Γ, F ) = 0 for all n > 0, i.e. sup n | Hn(Γ, F ) 6=0 for some F = 0.

2.7(a): Induced Γ-modules ZΓ ⊗ A are cohomologically trivial (as noted on pg148[1]) and hence haveprojective dimension ≤ 1 by Theorem VI.8.12[1].

2.7(b): If proj dimRM ≤ n, then Extn+1R (M,−) = 0 by Lemma VII.2.1[1]. For any direct summand

M ′ of M , we must then have Extn+1R (M ′,−) = 0 since Extn+1

R (−,−) commutes with direct sums. Thusproj dimRM

′ ≤ n by Lemma VII.2.1[1].

2.7(c): Suppose Γ is finite and M is a Γ-module in which |Γ| is invertible. It suffices to show thatM is a direct summand of an induced module ZΓ⊗A, for then proj dimZΓ(ZΓ⊗A) ≤ 1 by part(a) andhence proj dimZΓM ≤ 1 by part(b). Take A = M0, where M0 is the underlying abelian group of M . ByCorollary III.5.7[1] there is a Γ-module isomorphism ϕ : ZΓ⊗M → ZΓ⊗M0 given by g⊗m 7→ g⊗g−1m,where ZΓ⊗M has the diagonal Γ-action. The inclusion i : M → ZΓ⊗M given by m 7→

∑g∈Γ g ⊗m is

a Γ-module homomorphism because γ · i(m) = γ · (∑g g ⊗m) =

∑g γg ⊗ γm =

∑g g ⊗ γm = i(γ ·m).

The Γ-module map π : ZΓ ⊗M0 → M defined by g ⊗ m 7→ 1|Γ|gm is a Γ-splitting to ϕi [note: the

action on ZΓ⊗M0 is γ · (g ⊗m) = γg ⊗m]. Indeed, π[ϕi](m) = π[ϕ(∑g g ⊗m)] = π[

∑g g ⊗ g−1m] =

1|Γ|∑g gg

−1m = 1|Γ| |Γ|m = m = idM (m). Thus the injection ϕi : M → ZΓ ⊗M0 splits, and M is then

(by definition of a splitting homomorphism) a direct summand of ZΓ⊗M0 as a Γ-module.

4.1: If Z is finitely presented as a ZΓ-module then Z is finitely generated and every surjection P Z(with P finitely generated and projective) has a finitely generated kernel (Proposition VIII.4.1[1]). Inparticular, the augmentation map ε : ZΓ → Z has kernel I which then must be finitely generated (ob-viously noting that ZΓ is free of rank 1). Exercise I.2.1(d) then implies Γ is a finitely generated group.Conversely, suppose Γ is a finitely generated abelian group, so that Γ = F (S)/R is a group presentationfor Γ with |S| <∞. Then there is an exact sequence (ZΓ)|S| → ZΓ→ Z→ 0 by Exercise IV.2.4(d) andhence Z is finitely presented as a ZΓ-module by Proposition VIII.4.1[1].

6.1: Let Γ be of type FL and cd Γ = n. Then Γ is of type FP and hence of type FP∞ by Propo-sition VIII.6.1, and so there is a partial resolution Fm → · · · → F0 → Z → 0 with each Fi free of finiterank by Proposition VIII.4.3 (for all m ≥ 0). Due to its cohomological dimension, we can make a finiteprojective resolution 0 → P → Fn−1 → · · · → F0 → Z → 0 with each Fi free and P projective. ThenProposition VIII.6.5 implies that P is stably free, and so there is some free module F of finite rank suchthat P ⊕F is free. Take the free resolution 0→ F → F → 0→ · · · → 0 and consider its direct sum withthe finite projective resolution. This gives us a finite free resolution for Z over ZΓ of length n.

6.3: Let Γ be of type FP and cd Γ = n. Then we have a finite projective resolution Pn → · · · →P0 → Z → 0, and taking the Hom-dual we obtain the resolution for cohomology which behaves as· · · → HomZΓ(Pn,ZΓ)→ 0. Since Pn is a finitely generated projective module, so is its Hom-dual; thusHn(Γ,ZΓ) is a finitely generated Γ-module.

58

9 Chapter IX: Euler Characteristics

1.1: For a ring A, suppose there is a Z-valued function r on finitely generated projective A-modules,satisfying r(P ⊕Q) = r(P )+r(Q) and r(A) = 1 and r(P ) > 0 if P 6= 0. I claim that A is indecomposable,i.e. that A cannot be decomposed as the direct sum of two non-zero left ideals. Indeed, a decompositionA = I ⊕ J yields the equation 1 = r(I) + r(J) because I and J are projective A-modules (direct sum-mands of the free module A). Since both ideals are non-zero, r(I) ≥ 1 and r(J) ≥ 1, and this yields thedesired contradiction 1 ≥ 1 + 1 = 2.

2.1: Let P be a finitely generated projective (left) A-module and let P ∗ = HomA(P,A) be its dual.Then P ∗ is a right A-module and we have P ∗ ⊗A P ∼= HomA(P, P ) by Proposition I.8.3. Consider thediagram

P ∗ ⊗A P∼= //

ev$$

HomA(P, P )

trxx

T (A)

where ev(u⊗ x) = u(x) is the evaluation map. The isomorphism is given by u⊗ x 7→ [p 7→ u(p) · x], andon basis elements ei ∈ P this image homomorphism is ei 7→

∑j u(ei)rj · ej , where x =

∑j rjej . Thus

the composition is u⊗x 7→ tr[p 7→ u(p) ·x] =∑i u(ei)ri =

∑i ri · u(ei) =

∑i u(riei) = u(x) = ev(u⊗x),

and the diagram is commutative.

2.4: Let F be a finitely generated free module and e : F → F a projection operator of F onto adirect summand isomorphic to P (this is idempotent since e2 = e). Then e = i idP π, where i and πare the inclusion and projection maps between F and P , so tr(e) = tr(i idP π) = tr(idP ) = R(P ).Thus R(P ) is equal to the trace of an idempotent matrix defining P .

2.5: Let Γ be a group and ϕ : ZΓ→ Z the augmentation map, and let P be a finitely generated projectiveΓ-module. Then Proposition 2.3 implies that trZ(Z⊗Γ idP ) = T (ϕ)(trZΓ(idP )) = T (ϕ)(RZΓ(P )). As thisis an element of T (Z) = Z, it is immediate that trZ(Z ⊗Γ idP ) is precisely the Z-rank of PΓ = Z ⊗Γ P .Thus εΓ(P ) = T (ϕ)(RZΓ(P )).

2.6(a): Let Γ be a finite group. From the definition, trZΓ/Z : T (ZΓ) → T (Z) lifts to the mapZΓ → T (Z) = Z given by γ 7→ trZ(µγ), where µγ : ZΓ → ZΓ is right-multiplication by γ. Takingµγ as a matrix over Z, it is the identity matrix for γ = 1 and is a matrix with zeros on the diagonal for1 6= γ ∈ Γ (γ permutes the basis elements). Thus trZΓ/Z(1) = |Γ| and trZΓ/Z(γ) = 0 for 1 6= γ ∈ Γ.Consequently, there is a well-defined homomorphism τ : T (ZΓ) → Z such that τ(1) = 1 and τ(γ) = 0for 1 6= γ ∈ Γ, and one has trZΓ/Z = |Γ| · τ .

2.6(b): Applying Proposition 2.4 to α = idP and ϕ : Z → ZΓ, and following the same method asin the proof of Exercise 2.5, and using the result of part(a), we have that rkZ(P ) = |Γ| · τ(RZΓ(P )) forany finitely generated projective Γ-module P . Thus ρΓ(P ) = τ(RZΓ(P )).

2.7: The previous exercise shows that ρΓ(P ) ∈ Z, for Γ a finite group. It is obvious from the definition ofρ that ρΓ(P ) > 0 if P 6= 0 (it must be greater than or equal to 1/|Γ|). Now for two finitely generated pro-jective Γ-modules P and Q, they are necessarily free Z-modules and hence rkZ(P⊕Q) = rkZ(P )+rkZ(Q).Furthermore, ρΓ(ZΓ) = rkZ(ZΓ)/|Γ| = |Γ|/|Γ| = 1. Thus ZΓ is indecomposable by Exercise 1.1.

4.1: The proof of Theorem IX.4.4[1] used the assumption that Γ was finite in order to apply PropositionIX.4.1[1] by replacing Γ by the cyclic subgroup Γ′ = 〈γ〉. The proposition requires Γ′ to be of finite index(for all γ ∈ Γ), and this will hold in general if |Γ| <∞.

4.3(b): It is a fact from representation theory that a finitely generated kΓ-module is determined upto isomorphism by its character. By part(a), the character is in bijective correspondence with theHattori-Stallings rank. Thus two finitely generated kΓ-modules are isomorphic iff they have the same

59

Hattori-Stallings rank.

4.3(c): Take k = Q. If Γ is finite and P is a finitely generated projective ZΓ-module then Q ⊗Z Pis a finitely generated projective QΓ-module, hence a finitely generated projective ZΓ-module by Ex-ercise I.8.2. Then by Theorem 4.4 there is an integer r such that RΓ(Q ⊗Z P ) = r · [1]. But thisHattori-Stallings rank is precisely that of (QΓ)r, so Q⊗Z P is a free QΓ-module by part(b).

4.4: Taking ρΓ(P ) = RΓ(P )(1) as a definition, the result is precisely Proposition 4.1 applied to γ = 1.

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10 Additional Exercises

1: Find a counterexample to the statement MG∼= MG for a G-module M .

For an arbitrary group G we have (ZG)G ∼= Z ⊗ZG ZG ∼= Z. Alternatively, (ZG)G ∼= ZG/[I · ZG] =ZG/I ∼= Z (where the latter isomorphism follows from application of the 1st Isomorphism Theorem on theaugmentation map). If G is finite, then the norm element N exists, and the integer multiples zN (z ∈ Z)are the only elements of ZG fixed by all g ∈ G (assuming left-multiplication action), so (ZG)G = Z ·N[∼= Z] is the ideal generated by N . But if G is infinite then there is no norm element, and (ZG)G = 0.To prove this last statement, take any nonzero element x =

∑i rigi of ZG (since it’s a finite sum we

can assume 0 ≤ i ≤ n and all ri ∈ Z are nonzero) and consider the set S = g0, . . . , gn. To show thatx /∈ (ZG)G it suffices to show that there is at least one nontrivial g with gS 6= S. We can assume 1 /∈ S,otherwise for any g /∈ S we have g · 1 = g /∈ S. Suppose that gS = S ∀ g ∈ G. Then g0 · g0 = g2

0 ∈ S,and so through trivial induction we see that gi0 ∈ S (1 ≤ i ≤ n+ 1) and g0 · gn+1

0 = gn+20 /∈ S (otherwise

|S| > n+ 1), noting that gi0 6= 1 since 1 /∈ S. Thus we have arrived at a contradiction, and so the choicesG = Z and M = ZG suffice.

Another solution uses the choices G = Z2 = 〈x〉 and M = Z where G acts by x · n = −n. Thenx · n = n ⇒ −n = n ⇒ 2n = 0, so the largest quotient on which G acts trivially is ZZ2

= Z2, and2n = 0 only holds for n = 0 ∈ Z, hence ZZ2 = 0.

2: Let Z2 = 〈x〉 act on the additive complex numbers C by x · z = z∗, where z∗ = x − iy is thecomplex conjugate of z = x+ iy. [Note: CZ2 ∼= CZ2

∼= R]. Find H1(Z2,C).

For a derivation f : Z2 → C we have f(x0) = f(x2) = f(1) = 0 and f(x) determines f . Nowf(x2) = f(x) + x · f(x) = f(x) + f(x)∗ and so we must have f(x) = −f(x)∗ (i.e. a pure-imaginarynumber). Since iy = −(iy)∗ and 0 + iy = (iy)/2 + (iy)/2 = (iy)/2− (−iy)/2 = (iy)/2− (iy)∗/2, we havef(x) = 1

2f(x) − 12f(x)∗ = − 1

2f(x)∗ − x · [ 12f(x)] = x · [ 1

2f(x)∗] − [ 12f(x)∗]. Therefore, f is a principal

derivation and hence H1(Z2,C) = 0, using the result of Exercise III.1.2 above.

3: Let the multiplicative cyclic group C2k = 〈x〉 act on Z by x · n = (−1)kn. Calculate H1(C2k,Z)under this action.

For k = 2m even, C2k acts trivially on Z, so H1(C2k,Z) ∼= Hom(C2k,Z) = 0 where this equation followsfrom Exercise III.1.2 above. For k = 2m+1 odd, we start by viewing the 1-cocycles as Z1 = Der(C2k,Z)and the 1-coboundaries as B1 = PDer(C2k,Z). For f ∈ Z1, f(x2i) = f(x)+x·f(x2i−1) = f(x)−f(x2i−1)and f(x2i) = f(x2i−1) + x2i−1 · f(x) = f(x2i−1) + (−1)(2i−1)(2m+1)f(x) = f(x2i−1) − f(x), so 2f(x) =2f(x2i−1) ⇒ f(x) = f(x2i−1) and f(x2i) = 0. So Z1 consists of the derivations which are determinedby f(x) and satisfy the derived properties, hence Z1 ∼= Z. For f ∈ B1, f(xi) = xi · n− n = (−1)in− nwhich satisfies f(xi) = 0 ∀ i = 2j and f(xi) = −2n ∀ i = 2j + 1, and so B1 ∼= 2Z. ThusH1(C2k,Z) ∼= Z/2Z = Z2 when k is odd.Alternatively (for k odd), we know that H1(C2k,Z) = KerN where the map N : ZC2k

→ ZC2k is thenorm map induced by multiplication on Z by the norm element N . Now Nm = 1 · m + x · m + x2 ·m + · · · + x2k−1m = m − m + m − m + · · · + m − m = 0 and so NZ = 0 ⇒ KerN = ZC2k

. Sincexi · n = (−1)in, the action is trivial for all g ∈ C2k if −n = n, hence the largest quotient on which C2k

acts trivially is ZC2k= Z2.

4: Noting that the only nontrivial map ϕ : Zi = 〈t〉 → Zij = 〈s〉 is the canonical inclusion definedby t 7→ sj , show that the induced map under H2n−1 is the same inclusion. This is the corestriction map

corZijZi : H∗(Zi)→ H∗(Zij).

Considering the two periodic free resolutions of Z, there exists an augmentation-preserving chain map fbetween them by Theorem I.7.5[1] and we look at the commutative diagram in low dimensions:

61

Z[Zi]

f1

t−1// Z[Zi]

f0

ε // Z

idZ

// 0

Z[Zij ]s−1// Z[Zij ]

ε // Z // 0

The right square yields f0(1) = 1, and the left square yields (s− 1)f1(1) = f0(t− 1) = t · f0(1)− f0(1) =ϕ(t)1 − 1 = sj − 1 = (s − 1)(1 + s + · · · + sj−1) ⇒ f1(1) = 1 + s + · · · + sj−1. We assert that f isgiven by fn(1) = 1 for n even and fn(1) = 1 + s + · · · + sj−1 for n odd. Indeed, assuming inductivelythat the chain map is valid up to n, it suffices to check commutativity (1 + s + · · · + sij−1)fn+1(1) =fn(1 + t + · · · + ti−1) for n odd and (s − 1)fn+1(1) = fn(t − 1) for n even. For the latter case [n even]we have (s− 1)(1 + s+ · · ·+ sj−1) = (sj − 1) and fn(t− 1) = (sj − 1)fn(1) = sj − 1, so commutativityis satisfied. For the former case [n odd] we have fn(1 + t + · · · + ti−1) = (1 + sj + · · · + sij−j)fn(1) =(1+sj+· · ·+sij−j)(1+s+· · ·+sj−1) = 1+s+· · ·+sij−1 and (1+s+· · ·+sij−1)fn+1(1) = (1+s+· · ·+sij−1),so commutativity is satisfied.Using this chain map, and after moving to quotients, the cycle elements (for odd-dimensional homology)are mapped via ϕ∗(1) = 1 + 1 + · · ·+ 1 = j while the boundary elements are mapped via ϕ∗(1) = 1; thusthe result follows.

5: Let d : G→ A be a derivation. Prove the relation d(xn) = xn−1x−1 dx for x ∈ G.

For n = 0, d(1) = d(x0) = 1−1x−1dx = 0, and we argue by induction on n. Assuming the relation at

n = k holds, d(xk+1) = d(x · xk) = d(x) + x · d(xk) = [1 + xxk−1x−1 ]dx = [x−1+xk+1−x

x−1 ]dx = xk+1−1x−1 dx and

we are finished.Alternatively, we note that xn−1

x−1 = 1 + x + · · · + xn−2 + xn−1, so the relation immediately follows by

successive calculations d(xn) = d(x) + x · d(xn−1) = d(x) + x[d(x) + x · d(xn−2)] = d(x) + x · d(x) +x2[d(x) + x · d(xn−3)] = [1 + x+ x2]d(x) + x3 · d(xn−3), etc..

6: What information do we obtain about the homology of a group G by computing its homology withrational coefficients?

Assuming Q is an abelian group with trivial G-action, we can apply the result of Exercise III.1.2 to obtainthe short exact sequence 0→ Hn(G)⊗Q→ Hn(G,Q)→ TorZ1 (Hn−1(G),Q)→ 0. Since Q is torsion-freewe have the equality TorZ1 (Hn−1(G),Q) = 0 and hence the isomorphism Hn(G,Q) ∼= Hn(G) ⊗ Q. Now

A⊗Q = 0 for any torsion abelian group A because q⊗ a = |a|q|a| ⊗ a = q

|a| ⊗ |a|a = q|a| ⊗ 0 = 0. Therefore,

dimQ(Hn(G,Q)) = rkZ(HnG). Moreover, if Hn(G,Q) is nontrivial then HnG is torsion-free.

7: Let the multiplicative cyclic group Cn = 〈x〉 act on M =⊕n

j=1 Z2 by x · ai = ai+1 where ajgenerates the jth Z2-summand. Compute H1(Cn,M).

For f ∈ Z1 = Der(Cn,M) we have 0 = f(xn) = f(x) + xn−1 · f(x) ⇒ xn−1 · f(x) = −f(x) = f(x) [wecan drop the negative sign because the maximum order for elements is 2]. So f is determined by f(x) =(z1, . . . , zn) and we must have (z1, . . . , zn) = xn−1 · (z1, . . . , zn) = (z2, . . . , zn, z1) ⇒ z1 = z2 = · · · = zn.Thus f(x) is 0 or (1, 1, . . . , 1), and Z1 = Z2. But f(xi) = f(x) + xi−1 · f(x) = f(x) + f(x) = 2f(x) = 0,and f(x) = (1, 1, . . . , 1) = x · n− n where n ∈M is the element consisting of alternating 1’s and 0’s [thecase f(x) = 0 is trivial]. Thus f ∈ B1 = PDer(Cn,M), and so H1(Cn,M) = 0.

8: Let GLn(Z) act on Zn by left matrix multiplication, where we consider Zn as an n × 1 columnvector with integral entries. Compute the induced map ψ : H∗(GLn(Z),Zn) → H∗(GLn(Z),Zn) underthe action of −[δij ] on z ∈ Zn.

The anti-identity matrix m = −[δij ] is in the center Z(GLn(Z)) and so the conjugation action onGLn(Z) by m is the identity. Thus we can rewrite the map of the action m · z as (g 7→ mgm−1 =g , z 7→ m · z = −z) ∈ (GLn(Z),Zn). By Proposition III.8.1[1] this map induces the identity on therespective homology with coefficients, hence ψ = id∗. [But clearly ψ = −id∗ because it’s induced fromz 7→ −z ∈ Zn. Thus 2 · id∗ = 0 and H∗(GLn(Z),Zn) is all 2-torsion].

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9: The cyclic group Cm is a normal subgroup of the dihedral group Dm = Cm o C2 (of symmetriesof the regular m-gon). There is a C2-action on Cm = 〈σ〉 given by σ 7→ σ−1. Determine the action ofC2 on the homology H2i−1(Cm,Z), noting that there is an element g ∈ Dm such that gσg−1 = σ−1.

Letting c(g) : Cm → Cm be conjugation by g, we can apply Corollary III.8.2[1] to obtain the in-duced action of Dm/Cm ∼= C2 on H∗(Cm,Z) given by z 7→ c(g)∗z. It suffices to compute c(g)∗ onthe chain level, using the periodic free resolution P of Cm, and using the trivial action on Z. Us-ing the condition τ(hx) = [c(g)](h)τ(x) = h−1τ(x) on the chain map τ : P → P (for h ∈ Cm), weclaim that τ2i−1(x) = τ2i(x) = (−1)iσm−ix for i ∈ N and τ0(x) = x. Assuming this claim holds,the chain map P ⊗Cm Z → P ⊗Cm Z [in odd dimensions] is given by x ⊗ y 7→ (−1)iσm−ix ⊗ gy =(−1)iσm−ix ⊗ y = (−1)ix ⊗ σi−my = (−1)ix ⊗ y, and so c(g)∗ [hence the C2-action] is multiplicationby (−1)i on H2i−1(Cm,Z). It suffices to prove the claim. Seeing that Nτ2i(1) = τ2i−1(N) = Nτ2i−1(1)where N is the norm element, we can restrict our attention to τ2i−1 and use induction on i since(σ − 1)τ1(1) = (σ − 1)(−σm−1) = σm−1 − 1 = σ−1 − 1 = τ0(σ − 1). This chain map must satisfycommutativity (σ − 1)τ2i−1(1) = τ2(i−1)(σ − 1), and this is indeed the case because (σ − 1)τ2i−1(1) =(−1)i(σm−i+1 − σm−i) and τ2(i−1)(σ − 1) = (σ−1 − 1)(−1)i−1σm−i+1 = (−1)i(σm−i+1 − σm−i).

10: Assuming |G : H| < ∞, we define tr : MG → MH by tr(m) =∑g∈H\G gm, and we then define

the transfer map resGH : H∗(G,−)→ H∗(H,−) to be the unique extension of tr to a map of homologicalfunctors (using Theorem III.7.3[1]). Show that this agrees with the map defined by applying H∗(G,−)to the canonical injection M → CoindGHM

∼= IndGHM and using Shapiro’s lemma. Assume that we knowthis latter map is compatible with ∂.

By Theorem III.7.3[1] it suffices to verify that the latter map equals tr in dimension zero.We first want an explicit isomorphism for CoindGHM

∼= IndGHM , asserting that it is given by ψ(f) =∑g∈G/H g⊗f(g−1). Following the proof of Proposition III.5.9[1], there is an H-map ϕ0 : M → CoindGHM

given by [ϕ0(m)](g) = gm ∀ g ∈ H , 0 otherwise, and by the universal property of inductionthis extends to a G-map ϕ : IndGHM → CoindGHM given by ϕ(g ⊗ m) = gϕ0(m). Now we haveψϕ : g′ ⊗ m 7→ g′[ϕ0(m)] 7→

∑g∈G/H g ⊗ [ϕ0(m)](g−1g′) = g′′ ⊗ g′′−1g′m = g′′g′′−1g′ ⊗ m = g′ ⊗ m,

noting that there is one and only one g′′ ∈ G/H such that g′′−1g′ ∈ H (otherwise g−11 g′ = h1 and

g−12 g′ = h2 gives g1h1 = g2h2 which contradicts their coset representations). In the other direction we

have ϕψ : f 7→∑g∈G/H g⊗f(g−1) 7→

∑g∈G/H g[ϕ0(f(g−1))](g′) = [ϕ0(f(g′′−1))](g′g′′) = g′g′′f(g′′−1) =

f(g′g′′g′′−1) = f(g′) = f , also noting that there is only one g′′ ∈ G/H such that g′g′′ ∈ H. Thus ϕ = ψ−1

and ψ is the desired isomorphism.Applying the zeroth homology functor H0(G,−) to the aforementioned map M → CoindGHM → IndGHMgiven by m 7→ (g 7→ gm) 7→

∑g∈G/H g ⊗ g−1m, and using the isomorphism from Shapiro’s lemma,

we obtain the chain map x ⊗G m 7→∑g∈G/H x ⊗G (g ⊗H g−1m) =

∑g∈G/H xg ⊗G (1 ⊗H g−1m) ∼=∑

g∈G/H xg ⊗H g−1m =∑g∈G/H g

−1x⊗H g−1m =∑g∈H\G gx⊗H gm =

∑g∈H\G g · (x⊗H m). Using

the natural isomorphism H0(G,−) ∼= (−)G of Proposition III.6.1[1], this chain map yields the trace maptr described above.

11: Prove that Hn(G,M) ∼= TorGn−1(I,M) where I is the augmentation ideal of ZG.

We have the short exact sequence of G-modules, 0 → I → ZG ε→ Z → 0. By an analogue of The-orem 17.1.15[2] we have a long exact sequence of abelian groups · · · → TorGn (ZG,M) → TorGn (Z,M) →TorGn−1(I,M) → TorGn−1(ZG,M) → · · · . It is a fact that if P is R-projective then TorRn (P,B) = 0 for

any R-module B (n ≥ 1). Therefore, TorGn (ZG,M) = 0 since ZG is a free [hence projective] ZG-module,and so we obtain the isomorphisms TorGn (Z,M) → TorGn−1(I,M). Since Hn(G,−) = TorGn (Z,−), theresult follows.

12: Given |G : H| < ∞ and z ∈ H(G,M) where H(−,−) is either H∗ or H∗, show that corGHresGHz =|G : H|z.

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Referring to Exercise AE.10 above, we have the map M → CoindGHM → IndGHM which yields thechain map x ⊗G m 7→

∑g∈H\G gx ⊗H gm, and this induces the transfer map resGH in homology. Com-

posing this with the corestriction map on the chain level x ⊗H m 7→ x ⊗G m, we obtain the chain mapx⊗Gm 7→

∑g∈H\G gx⊗G gm =

∑g∈H\G x⊗Gm = |G : H|(x⊗Gm) which induces corGHresGH as multi-

plication by |G : H|.On the other hand, we have the chain map Hom(F,M)H → Hom(F,M)G given by f 7→

∑g∈G/H [x 7→

gf(g−1x)] which induces the transfer map corGH in cohomology; G acts diagonally. Composing this withthe restriction map on the chain level, we obtain the chain map f 7→ f 7→

∑g∈G/H [x 7→ gf(g−1x)] =∑

g∈G/H [x 7→ f(x)] = |G : H|[x 7→ f(x)], noting that the domain element f is a G-invariant homomor-

phism. This induces the aforementioned map corGHresGH .

13: Give another proof that H(G,⊕m

i Mi) ∼=⊕m

i H(G,Mi) where H(−,−) is either H∗ or H∗.

Applying Proposition III.6.1[1] to the short exact sequence of G-modules 0→M1α→M1⊕M2

β→M2 → 0,we obtain the standard long exact sequence in (co)homology. Since α is an injection onto a direct sum-mand, the induced map α∗ under the covariant functor H∗(G,−) is an injection (refer to Exercise II.7.3);similarly, the induced map β∗ = H∗(G, β) is also an injection. Thus the long exact sequence breaks upinto short exact sequences 0 → Hn(G,M1) → Hn(G,M1 ⊕M2) → Hn(G,M2) → 0 (similar for coho-mology). These are split exact sequences because γ∗α∗ = id∗ and Γ∗β∗ = id∗, where γ is the projectionM1 ⊕M2 → M1 and Γ is the inclusion M2 → M1 ⊕M2. The result now follows by trivial induction onm.

14: Prove that if the G-module M has exponent p [prime] then Hn(G,M) and Hn(G,M) are Zp-vector spaces.

Showing that these groups are the specified modules is equivalent to showing that they are annihi-lated by p (i.e. have exponent dividing p). Considering homology, a chain z ∈ Cn(G,M) is a finite sumof terms of the form m ⊗ [g1| · · · |gn], so pz =

∑p(m ⊗ [g1| · · · |gn]) =

∑(pm) ⊗ [g1| · · · |gn] =

∑0 = 0.

Thus pHn(G,M) = 0 ∀ n ≥ 0. Considering cohomology, a cochain f ∈ C0(G,M) is an element of Mand hence pf = 0 trivially. If n ≥ 1 then f ∈ Cn(G,M) is a function Gn → M , so pf is a functionGn →M → pM = 0 and hence pf = 0. Thus pHn(G,M) = 0 ∀ n ≥ 0.

15: If G is a finite group, show that Hn(G,M) is a Z|G|-module for n > 0. Thus we have the pri-mary decomposition Hn(G,M) =

⊕pHn(G,M)(p) where p ranges over the primes dividing |G|.

By Proposition III.9.5[1], corG1resG1z = |G : 1|z = |G|z where z ∈ H∗(G,M). But this composition

factors through H∗(1,M) which is trivial in positive dimensions. Thus corG1resG1z = 0 ⇒ |G|z = 0,

and so Hn(G,M) is annihilated by |G| for n > 0.

16: Show that Hn(G,M) = 0 ∀ n > 0 if M is a finite abelian group and G is finite with rela-tively prime order, gcd(|G|, |M |) = 1. Furthermore, show that Hn(G,M) is a finite group (for n > 0) ifG is finite and M is a finitely generated abelian group.

By Exercises AE.14+15 above, Hn(G,M) is annihilated by both |G| and |M |, and hence must be anni-hilated by a common factor. But gcd(|G|, |M |) = 1, so the common factor is 1 and Hn(G,M) = 0 forn > 0.By Exercise AE.15 above (assuming n > 0), |G|Hn(G,M) = 0 and so Hn(G,M) is all torsion. Since|G| <∞, Fn is a free G-module of finite rank r = |G|n (F is the bar resolution of Z over ZG), where wenote that a G-basis for Fn is the (n+1)-tuples whose first element is 1. Thus Fn⊗GM ∼= (

⊕ri ZG)⊗GM ∼=⊕r

i (ZG⊗GM) ∼=⊕r

i M is finitely generated since M is finitely generated, which implies that Hn(G,M)is also finitely generated. A finitely generated torsion group is finite, so the result follows.

17: Why does resGH induce a monomorphism Hn(G,M)(p) → Hn(H,M), where H is the Sylow p-subgroup of G?

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Consider an element z ∈ Hn(G,M)(p) which lies in the kernel of resGH . Composing this with the core-striction map gives you multiplication by |G : H| by Proposition III.9.5[1], and so |G : H|z = 0. But theorder of z is |z| = pα and p does not divide |G : H|. Since pα and |G : H| both annihilate z, there is acommon factor between them which also annihilates z, so z = 0 (the only common factor is 1). ThusresGH is a monomorphism when restricted to the p-primary component.

18: Compute the homology group H1(G) for any nonabelian simple group G.

The commutator group [G,G] is either 0 or G because the commutator group is a normal subgroupof G and a normal subgroup of a simple group must be either the trivial group or G itself. Since G isnonabelian, we cannot have [G,G] = 0, and so [G,G] = G (i.e. G is perfect). Thus H1(G) ∼= G/[G,G] =G/G = 0.

In particular, H1(An) = 0 ∀ n ≥ 5 where An is the alternating group on n letters (subgroup of Snwith index 2), by Theorem 4.6.24[2].

19: Show that I/I2 ∼= Gab where I is the augmentation ideal of ZG.

Apply Proposition III.6.1[1] to the short exact sequence 0 → I → ZG → Z → 0 to obtain the ex-act sequence H1(G,ZG) → H1(G) → IG → (ZG)G → Z → 0 in low dimensions. The latter map is anisomorphism (by the First Isomorphism Theorem) because the map is surjective and (ZG)G ∼= Z. ByProposition III.6.1[1], H1(G,ZG) = 0 because ZG is free [hence projective]. Thus H1(G) ∼= IG = I/I2

by exactness, and we know that H1(G) ∼= Gab, so the result follows.An explicit isomorphism is given by g[G,G] 7→ (g − 1) + I2.

20: Find a module M with trivial action such that H1(D2n ,M) is nonzero, where D2n is the dihe-dral group with respect to a regular 2n−1-gon.

A presentation for this group is D2n = 〈α, β | α2 = β2 = (αβ)2n−1

= 1〉. By the result of ExerciseIII.1.2, H1(D2n ,M) = Hom(Dab

2n ,M). Now D2n quotiented by [D2n , D2n ] forces the trivial relation 1 =

(αβ)2n−1

= α2n−1

β2n−1

= 1 ·1 = 1, so Dab2n = Z2⊕Z2. Thus H1(D2n ,M) = Hom(Z2,M)⊕Hom(Z2,M),

which is nonzero if M = Z2, giving H1(D2n ,Z2) ∼= Z2 ⊕ Z2.

21: Prove that a finitely generated abelian group G is cyclic if and only if H2(G,Z) = 0.

If G is a cyclic group Zn, then H2(Zn,Z) = 0 as explained on pg35[1].If H2(G,Z) = 0 then Hopf’s formula (Theorem II.5.3[1]) gives R∩ [F, F ] ⊆ [F,R], where F/R is a presen-tation for G. Since G is abelian we have 0 = [G,G] = [F/R,F/R] = [F, F ]R/R which implies [F, F ] ⊆ R.Alternatively, since G = F/R is abelian, [F, F ] ⊆ R by Proposition 5.4.7[2]. Thus [F, F ] = [F,R], and|F | < ∞ since G is finitely generated. If |F | = 1 then R = xn for some n ≥ 0, in which case G isobviously cyclic.It seems we’re stuck in terms of extracting any more information, but alternatively we can refer toTheorem V.6.3[1] which says H2(G) is isomorphic to the second exterior power

∧2G for any abelian

group. Thus it suffices to show that G is cyclic if G ⊗ G ∼= 〈g ⊗ g〉 ∼= G. Now if rkZ(G) = n thenrkZ(G⊗G) = n2, so n2 = n gives n = 1 or n = 0. Also, Zi ⊗ Zj ∼= Zgcd(i,j), so G must contain at mostone Zm-summand (for each m) because the tensor product contains at least the squared-amount of thosesummands. Assume that G is not cyclic; then all m’s must be relatively prime (otherwise the tensorproduct will contain additional summands not in G due to greatest common divisors). Thus G ∼= Z⊕Zn(for some n ≥ 2) since Zi⊕Zj ∼= Zij iff gcd(i, j) = 1 [note that G must have the Z-summand in order tonot be cyclic]. But (Z⊕Zn)⊗ (Z⊕Zn) ∼= Z⊕Zn⊕Zn⊕Zn which is not isomorphic to G, a contradiction(hence G is cyclic).Aside: A finitely generated abelian group G = F/R is cyclic iff [F, F ] = [F,R].

22: Let Zm = 〈t〉, Zm2 = 〈s〉, and define a Zm-action on Zm2 by t · s = sm+1. Compute the re-sulting Zm-module structure on Hj(Zm2), and then compute Hi(Zm, Hj(Zm2)) where m is an odd prime.

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First we must compute the change-of-rings map in integral homology for the map Za = 〈x〉 → Zb = 〈y〉,x 7→ yr, where b|ra. Mimicking the solution to Exercise AE.4, the induced map in odd-dimensionalhomology is easily seen to be multiplication by r. Applying this result to the case where r = m+ 1 anda = b = m2, the Zm-action on Hj(Zm2) is multiplication by p(m+1) for tp. If j is even, then the homolo-gies in question are trivial. By a result on pg58[1], Hodd(Zm,Zm2) = Coker(N : (Zm2)Zm → (Zm2)Zm)and Heven(Zm,Zm2) = Ker(N : (Zm2)Zm → (Zm2)Zm). Since m is an odd prime, the only nontrivialproper subgroup/quotient of Zm2 is Zm which is not fixed by the Zm-action, so the norm map is 0→ 0and thus Hi(Zm, Hj(Zm2)) = 0 ∀ nonzero i, j.

23: Prove that if Q is injective then E(G,Q) is trivial.

Any short exact sequence 0 → Q → E → G → 0 splits for Q injective, so there is only one equiva-lence class of group extensions (namely, the split extension) and E(G,Q) = 0. Alternatively, PropositionIII.6.1[1] and Theorem IV.3.12[1] imply E(G,Q) ∼= H2(G,Q) = 0 because Q is injective.

24: Prove that H2(A5,Z2) 6= 0.

We have the central group extension 1 → Z2 → SL2(F5) → A5 → 1 as in Exercise I.5.7(b). It suf-fices to show that E = Z2 o A5 is not isomorphic to SL2(F5), for then the above extension is nonsplitand hence H2(A5,Z2) ∼= E(A5,Z2) 6= 0 by Theorem IV.3.12[1]. First note that Z2 must have the trivialA5-action, so E = Z2 × A5. Now it is a fact that SL2(F5) is perfect, while [E,E] ⊆ A5 by Proposition5.4.7[2] because E/A5

∼= Z2 is abelian (hence E is not perfect). Thus H1(E) 6= 0 = H1(SL2(F5)) and soZ2 oA5 SL2(F5).

25: Let A = Z2×Z2 and let Aut(A) ∼= S3 act on A in the natural fashion. Prove that H1(S3,Z2×Z2) = 0.

In the semi-direct product E = AoS3 = Hol(A) [called the holomorph of A] we have a Sylow 3-subgroupP ∼= Z3 by Sylow’s Theorem, where |E| = 24 = 23 · 3. By Sylow’s Theorem, n3 | 8 and n3 ≡ 1 mod 3,where n3 is the number of Sylow 3-subgroups of E. Thus n3 is either 1 or 4, and this implies |NE(P )|is either 24 or 6 by the fact n3 = |E : NE(P )|. But we can exhibit at least two such subgroups [notingthat S3

∼= Z3 o Z2], namely P1 = (0 × 0) o (Z3 o 0) and P2 = (0, 0, 0, 0), (1, 0, 1, 0), (1, 0, 2, 0). Thus|NE(P )| = 6 which corresponds to n3 = 4, and S3 is then the normalizer of the Sylow 3-subgroupP = 0oZ3 of E because P / S3 and |S3| = 6. Given a complement G to A in E [a group G is a comple-ment to A in E if E = AoG], one Sylow 3-subgroup is Q /G and hence P = eQe−1 for some e ∈ E bySylow’s Theorem. Now S3 = NE(P ) = NE(eQe−1) = eNE(Q)e−1 = eGe−1, and so all complements areconjugate. Noting that the conjugacy classes of complements to A in E are the A-conjugacy classes ofsplittings of E, we have H1(S3,Z2×Z2) = 0 by Proposition IV.2.3[1] because there is only one conjugacyclass.

26: In the proof of Theorem IV.3.12[1], all extensions were assumed to have normalized sections. Explainwhy this simplification does not affect the result of the theorem.

Given a factor set (2-cocycle) f : G × G → A, we assert that this lies in the same cohomologyclass as a normalized factor set (2-cocycle). Let δ1c be the coboundary of the constant functionc : G → A defined by c(g) = f(1, 1). It suffices to show that F = f − δ1c is a normalized factorset, for then it belongs to the same cohomology class as the arbitrary f [it differs by a coboundary]. NowF (1, 1) = f(1, 1)−[δ1c](1, 1) = f(1, 1)−[c(1)+1·c(1)−c(1·1)] = f(1, 1)−f(1, 1) = 0 and so F : G×G→ Ais normalized. It is indeed a 2-cocycle (factor set) because gF (h, k) − F (gh, k) + F (g, hk) − F (g, h) =[gf(h, k)− f(gh, k) + f(g, hk)− f(g, h)]− g[c(h) + hc(k)− c(hk)]− [c(gh) + ghc(k)− c(ghk)] + [c(g) +gc(hk)− c(ghk)]− [c(g)+gc(h)− c(gh)] = 0−gc(h)−ghc(k)+gc(hk)+ c(gh)+ghc(k)− c(ghk)− c(g)−gc(hk) + c(ghk) + c(g) + gc(h) − c(gh) = [−gc(h) + gc(h)] + [−ghc(k) + ghc(k)] + [gc(hk) − gc(hk)] +[c(gh)− c(gh)] + [−c(ghk) + c(ghk)] + [−c(g) + c(g)] = 0 + 0 + 0 + 0 + 0 + 0 = 0.Since the bijection in the theorem is dependent on the cohomology class, and every class has a normalizedfactor set (hence normalized section), we are able to restrict our attention to those sections satisfyingthe normalization condition.

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27: Show that H2(F,A) = 0 for F free by appealing to group extensions.

For any group extension 0→ A→ Eπ→ F → 1 define the set map S → E by s 7→ s, where s ∈ π−1(s) is

a lifting of s ∈ F = F (S). Then by the universal mapping property of free groups, this set map extendsuniquely to a homomorphism ϕ : F → E which satisfies πϕ = idF by construction. Thus the extensionsplits, and by Theorem IV.3.12[1] we have H2(F,A) ∼= E(F,A) = 0.Note that this also follows from direct computation using the free resolution 0 → I → ZF → Z → 0,where the augmentation ideal I of ZF is free by Exercise IV.2.3(b). Indeed, the coboundary map isδn : HomF (0, A)→ HomF (0, A) for all n > 1 and hence Hn(F,A) = 0 ∀ n > 1.

28: Compute H2(Q8,Z2), and determine the number of group extensions of Z2 by Q8. [These twoproblems are independent of each other].

The quaternion group Q8 is a non-abelian group of order 8 with presentation 〈x, y | x4 = 1, x2 =y2 = (xy)2〉. Its center is C := Z(Q8) ∼= Z2 which is also its commutator subgroup, and Inn(Q8) =Q8/Z(Q8) ∼= Z2 × Z2. Also, Aut(Q8) ∼= S4 and hence Out(Q8) = Aut(Q8)/Inn(Q8) ∼= S3. As explainedon pg102+104[1], Q8 is an S4-crossed module via the canonical map Q8 → Aut(Q8), and such an exten-sion gives rise to a homomorphism ψ : Z2 → Out(Q8) ∼= S3. By Theorem IV.6.6[1] the set E(Z2, Q8, ψ)of equivalence classes of extensions giving rise to ψ is either empty or in bijective correspondence withH2(Z2, C), where C is a Z2-module via ψ [note: it is a fact that the center of a group is a characteristicsubgroup, so Aut(Q8) acts naturally on C and hence Out(Q8) acts on C because any inner automorphismleaves C fixed]. Now the only possible action on C ∼= Z2 is the trivial one (giving CZ2 = C = CZ2), soH2(Z2, C) = CokerN = C/NC = C/0 ∼= Z2. There are two possible choices of ψ, namely, the trivial mapand the injection Z2 → S3 = Z3oZ2. If ψ is the trivial map then it automatically lifts to the trivial ho-momorphism Z2 → S4 and we have a direct product extension E ∼= Q8×Z2. For the injective ψ we havethe semi-direct product Q8oZ2 where Z2 switches the generators of Q8. Thus E(Z2, Q8, ψ) is nonempty,and Theorem IV.6.6[1] implies there are a total of 4 group extensions 1→ Q8 → E → Z2 → 1.To compute the second cohomology group of the group Q8 with coefficients in Z2, embed Z2 in theH∗-acyclic module Hom(ZQ8,Z2) by z 7→ (q 7→ qz), and note that Coker(Z2 → Hom(ZQ8,Z2)) = Z2

because the evaluation map Hom(ZQ8,Z2) → Z2 given by f 7→ f(2) composes with the embeddingto give the trivial map z 7→ (q 7→ qz) 7→ 2z = 0. The dimension-shifting argument then impliesH2(Q8,Z2) ∼= H1(Q8,Z2), and by Exercise III.1.2 we have H1(Q8,Z2) ∼= Hom(H1(Q8),Z2). NowH1(Q8) = Q8/[Q8, Q8] = Q8/C ∼= Z2 × Z2 and so Hom(H1(Q8),Z2) ∼= Hom(Z2,Z2) ⊕ Hom(Z2,Z2) ∼=Z2 × Z2. Therefore, H2(Q8,Z2) ∼= H1(Q8,Z2) ∼= H1(Q8) ∼= Z2

2 ≡ Z2 × Z2.

29: Let p be a prime, let A be an abelian normal p-subgroup of a finite group G, and let P be aSylow p-subgroup of G. Prove that G is a split extension of G/A by A iff P is a split extension of P/Aby A [Note: it is a fact that a normal p-subgroup is contained in every Sylow p-subgroup, so A ⊆ P ].This result is known as Gaschutz’s Theorem.

Suppose G splits over A, so that G ∼= AoG/A. Then the subgroup Ao P/A is a Sylow p-subgroup ofG and hence A o P/A ∼= P by Sylow’s Theorem, so P also splits over A. Conversely, suppose P splitsover A (i.e. P ∼= AoP/A). Note that P/A = Sylp(G/A) and multiplication by |G/A : P/A| = |G : P | isan automorphism of A [hence of H2(G/A,A)] since |A| is relatively prime to |G : P |. The composition

H2(G/A,A)res→ H2(P/A,A)

cor→ H2(G/A,A) is then an isomorphism by Proposition III.9.5[1] because it ismultiplication by |G : P |. In particular, the restriction homomorphism res : H2(G/A,A)→ H2(P/A,A)is injective, so the only element of H2(G/A,A) which corresponds to the trivial element [split extension]of H2(P/A,A) is the trivial element [split extension]. Thus G splits over A, and the proof is complete.

30: The Schur multiplier of a finite group G is defined as H2(G,Z) ∼= H2(G,C∗) where the multiplica-tive group C∗ = C−0 is a trivial G-module. Prove that the Schur multiplier (of a finite group) is finite.

Since |G| < ∞ and Z is finitely generated (as an abelian group), H2(G,Z) is a finite group by Ex-ercise AE.16.Alternatively, we shall show that every cohomology class contains a cocycle whose values lie in the nth

67

roots of unity 〈ζ〉 ∼= Zn (where n = |G| and ζ = e2πi/n), for then there are only finitely many func-

tions/cocycles f : G2 → 〈ζ〉 and |H2(G,C∗)| ≤ nn2

< ∞. From the exact sequence 0 → Zn → C∗ n→C∗ → 0 we obtain a long exact sequence → H1(G,C∗) → H1(G,C∗) → H2(G,Zn) → H2(G,C∗) →H2(G,C∗)→ by Proposition III.6.1[1]. But the nth-power map on C∗ induces the n-multiplication mapon H2(G,C∗) which is the zero map since n annihilates H2(G,C∗) by Corollary III.10.2[1], so the abovelong exact sequence gives us a surjection H2(G,Zn)→ H2(G,C∗)→ 0. Now H2(G,Zn) is finite by Exer-cise AE.16, so H2(G,C∗) is necessarily finite (by the 1st Isomorphism Theorem) and the proof is complete.

31: Show that ZG⊗ ZG′ ∼= Z[G×G′] as (G×G′)-modules.

The map ZG×ZG′ → Z[G×G′] defined by (zg, z′g′) 7→ zz′(g, g′) is obviously a Z-balanced map and hencegives a rise to a unique group homomorphism ϕ : ZG⊗ZG′ → Z[G×G′] by Theorem 10.4.10[2]. The ob-vious group homomorphism Z[G×G′]→ ZG⊗ZG′ defined by z(g, g′) 7→ z(g⊗g′) = (zg⊗g′) = (g⊗zg′)is the inverse of ϕ because z(g, g′) 7→ (zg ⊗ g′) 7→ z1(g, g′) = z(g, g′) and (zg ⊗ z′g′) 7→ zz′(g, g′) 7→zz′(g ⊗ g′) = (zg ⊗ z′g′). Thus ϕ is a group isomorphism, and it is a (G × G′)-module isomorphismbecause ϕ[(h, h′) · (zg ⊗ z′g′)] = ϕ(zhg ⊗ z′h′g′) = zz′(hg, h′g′) = zz′(h, h′)(g, g′) = (h, h′) · ϕ(zg ⊗ z′g′)where (h, h′) ∈ G×G′.

32: Suppose u1 ∈ HpG, u2 ∈ HqG, v1 ∈ HrG, and v2 ∈ HsG. Prove that (u1 × v1) ` (u2 × v2) =(−1)qr(u1 ` u2)× (v1 ` v2) in Hp+q+r+s(G,Z).

Note that u ` v := d∗(u × v) where d : G → G × G is the diagonal map. Let D : G × G → G4 be theanalogous diagonal map, and let P : G4 → G4 be the permutation (g1, g2, g3, g4) 7→ (g1, g3, g2, g4). ThenD = P(d×d), and we obtain (u1×v1) ` (u2×v2) = D∗(u1×v1×u2×v2) = (d×d)∗[P ∗(u1×v1×u2×v2)] =(−1)qr(d × d)∗[u1 × u2 × v1 × v2] = (−1)qrd∗(u1 × u2) × d∗(v1 × v2) = (−1)qr(u1 ` u2) × (v1 ` v2).Another way is to perform the same calculation using u × v := p∗1(u) ` p∗2(v) and the fact thatp∗(u ` v) = p∗(u) ` p∗(v), where p1 is the projection G × G → G × 1 = G and p2 is the pro-jection G×G→ 1 ×G = G.

33: For the cap product, state the property of naturality with respect to group homomorphismsα : G→ H. Also, provide the existence of an identity element for the cap product.

Checking definitions, we have u a α∗z = α∗(α∗u a z) which is associated to the “commutative” di-

agram

Hp(G,M)⊗Hq(G,N)

α∗

a// Hq−p(G,M ⊗N)

α∗

Hp(H,M)⊗Hq(H,N)

α∗

DD

a// Hq−p(H,M ⊗N)

There is a left-identity element 1 ∈ H0(G,Z) = Z which is represented by the augmentation map ε,regarded as a 0-cocycle in HomG(F,Z). Let F be the standard resolution, let z ∈ H∗(G,M) with repre-sentation z = (g0, . . . , gi)⊗m, and take the diagonal approximation ∆ to be the Alexander-Whitney map.

Then ε⊗ z maps under a to∑ip=0(−1)degε·p(g0, . . . , gp)⊗ ε(gp, . . . , gi)⊗n =

∑0 + (−1)0·i(g0, . . . , gi)⊗

1 ⊗ n = z, as explained on pg113[1]. Thus the element satisfies 1 a z = z for all z ∈ H∗(G,M), wherewe make the obvious identification Z⊗M = M of coefficient modules.

34: Prove that a finitely generated projective Z-module M is a finitely generated free Z-module; thusthe two are actually equivalent (since free modules are projective).

By the Fundamental Theorem (Theorem 12.1.5[2]) M has the decomposition M ∼= Zr ⊕Zn1⊕ · · · ⊕Zni ,

where we note that Z is a Principal Ideal Domain. Since M is projective, all of its direct summandsmust be projective. Now Zr is free, hence projective. But Zn is not Z-projective because if it were thenit would be a direct summand of a free Z-module F , and F would then have elements of finite order (acontradiction); alternatively we could note that applying the functor Hom(Zn,−) to the exact sequence

0→ Z n→ Z→ Zn → 0 yields the sequence 0→ 0→ 0→ Zn → 0 which is not exact. Thus M ∼= Zr for

68

some r ∈ N ∪ 0 and hence M is Z-free of finite rank.

35: Let G = PSL2(Z) and let A be a G-module. Show that for every q ≥ 2 and for every x ∈ Hq(G,A),we have 6x = 0.

The modular group is the group of Mobius transformations T (z) = az+bcz+d in the complex plane such

that a, b, c, d ∈ Z with ad − bc = 1. It is isomorphic to G via the map T 7→(a bc d

), and G is called the

projective special linear group [the quotient of SL2(Z) by Z(SL2(Z)) ∼= Z2]. Now it is a fact that Gcontains a free subgroup H of index |G : H| = 6 [this fact can be found in the paper The Number ofSubgroups of Given Index in the Modular Group by W. Stothers]. By Exercise AE.27, Hq(H,A) = 0 forall q ≥ 2. Thus we can apply Proposition III.10.1[1] which states Hq(G,A) is annihilated by |G : H|, i.e.6x = 0 for all x ∈ Hq(G,A) with q ≥ 2.

36: Let G be a finite cyclic group and let M be a G-module. The Herbrand quotient is defined tobe h(M) = |H2(G,M)|/|H1(G,M)|, assuming both cohomology groups are finite. Show that h(Z) = |G|where Z has the trivial G-action, and show that h(M) = 1 for M finite.

We know that H2(G,M) ∼= MG/NM and H1(G,M) = NM/IM , where N is the norm element andI ≡ 〈σ− 1〉 is the augmentation ideal of G = 〈σ〉 and NM := Ker(N : M →M). For M = Z with trivialaction we have H2(G,M) ∼= Z|G| and H1(G,M) = 0, so h(Z) = |ZG|/|0| = |G|/1 = |G|. If M is finitethen h(M) = (|MG| · |IM |)/(|NM | · |NM |) = |MG| · |(σ−1)M |/|M |, where we note that M/NM ∼= NM .

But the kernel K of the surjective map Mσ−1→ (σ − 1)M is equal to MG because m ∈ MG maps to

σm −m = m −m = 0 and if m ∈ K then σ|G|−1m = σ|G|−2m = · · · = σm = m, i.e. m ∈ MG. Thus|M |/|MG| = |(σ − 1)M | and so h(M) = |M |/|M | = 1.

37: Given the short exact sequence of G-modules 0→ A→ B → C → 0 with G finite cyclic, show thath(B) = h(A)h(C) where we assume the cohomology groups for A and C [hence B] are finite. Here h isthe Herbrand quotient (defined in the previous exercise).

Consider the long exact cohomology sequence H0(G,C)δ0→ H1(G,A)

ϕ→ H1(G,B)→ · · · → H2(G,B)φ→

H2(G,C)δ2→ H3(G,A). It is a fact (Exercise V.3.3(b)) that cupping this sequence with the generator

of H2(G,Z) gives an isomorphism of that sequence onto itself (raising dimensions by 2), and hencegives the equality δ0 = δ2 which implies Kerϕ ∼= Cokerφ. We can break the above sequence and ob-

tain an exact sequence 0 → Kerϕ → H1(G,A)ϕ→ · · · φ→ H2(G,C) → Cokerφ → 0. I claim that

if 0 → H1 → · · · → Hm → 0 is an exact sequence of finite abelian groups, then∏mi=1 |Hi|(−1)i =

1. Assuming this claim holds, and noting that the cohomology groups are finite abelian, we have

1 = |Cokerφ||Kerϕ|

|H1(G,A)||H1(G,B)|

|H1(G,C)||H2(G,A)|

|H2(G,B)||H2(G,C)| = 1 · 1

h(A)1

h(C)h(B) which implies h(B) = h(A)h(C) as desired.

It suffices to prove the claim. The case m = 2 is trivial since the sequence yields the isomorphismH1∼= H2 and hence |H1|/|H2| = 1, so we proceed by induction on m > 1. From the exact sequence

0 → H1 → · · · → Hm−1φ→ Hm

ϕ→ Hm+1 → 0 we obtain the exact sequence 0 → H1 → · · ·Hm−2 →Hm−1

φ→ Imφ→ 0. Applying the inductive hypothesis we have∏m−1i=1 |Hi|(−1)i · |Imφ|(−1)m = 1, and by

exactness of the original sequence we see that |Imφ| = |Kerϕ|. But |Hm|/|Kerϕ| = |Imϕ| = |Hm+1|, so

|Kerϕ| = |Hm|/|Hm+1| and hence 1 =∏m−1i=1 |Hi|(−1)i ·(|Hm|/|Hm+1|)(−1)m =

∏m+1i=1 |Hi|(−1)i as desired.

38: If G and H are abelian groups with isomorphic group rings ZG ∼= ZH, show that G ∼= H.

The homology groups of G and H are independent of the choice of resolution up to canonical iso-morphism, and the groups are defined by HiG = Hi(FG) where F is a projective resolution of Z over ZG(similary for H). Since ZG ∼= ZF , the G-modules Fi can be regarded as H-modules via restriction ofscalars and hence the projective resolution F for the homology of G can also be used for the homologyof H. We have FG ∼= FH since Z ⊗ZG F ∼= Z ⊗ZH F by the obvious map 1 ⊗ f 7→ 1 ⊗ f [using theisomorphism ϕ : ZG → ZH we have 1 ⊗ f = 1 ⊗ gf 7→ 1 ⊗ ϕ(g)f = 1 ⊗ f ], and so Hi(G) ∼= Hi(H)for all i. In particular, G/[G,G] ∼= H1(G) ∼= H1(H) ∼= H/[H,H]. Since G and H are abelian groups,

69

[G,G] = 0 = [H,H] and hence G ∼= H.

39: Let F be a field and G a finite group of order |G| > 1. Show that the group algebra FG haszero divisors (hence is not a field) and show that the augmentation ideal I is a maximal ideal of FG.

Noting that FG/I ∼= F from the exact sequence 0 → I → FGε→ F → 0, I is a maximal ideal by

Proposition 7.4.12[2] since F is a field [in a commutative ring R, an ideal M is maximal iff R/M is afield]. Now take g ∈ G such that gm = 1 with m > 1. Then (1−g)(1+g+· · ·+gm−1) = 1−gm = 1−1 = 0and hence 1− g is a zero divisor.

40: Regarding Z2 as a module over the ring Z4, construct a resolution of Z2 by free modules overZ4 and use this to show that ExtnZ4

(Z2,Z2) is nonzero for all n.

The 2-multiplication map Z42→ Z4 has kernel 0, 2 and image 0, 2, and the projection Z4 Z2

has kernel 0, 2 and image 0, 1. Thus we have a free resolution · · · → Z42→ Z4

2→ Z4 Z2 → 0.Since Hom(Z4,Z2) = Z2, the 2-multiplication maps become trivial maps, and we obtain the sequence

· · · ← Z20← Z2

0← Z2 after applying Hom(−,Z2) to our free resolution. Thus ExtnZ4(Z2,Z2) ∼= Z2 6= 0

for all n.

41: Let H be a subgroup of G of finite index, and let 0 → A′i→ A

j→ A′′ → 0 be an S.E.S. of G-modules. Show that the connecting homomorphisms are consistent with corestriction, i.e. the followingdiagram commutes.

Hn(G,A′′)δ // Hn+1(G,A′)

Hn(H,A′′)δ //

cor

OO

Hn+1(H,A′)

cor

OO

Let F be the resolution associated to G and let F ′ be the resolution associated to H, and let ∂∗

denote the [cochain complex] coboundary map. Consider the exact sequence of cochain complexes

0 → HomH(F ′n, A′)

iH→ HomH(F ′n, A)jH→ HomH(F ′n, A

′′) → 0, with similar notation for the sequenceassociated to G. The connecting map δ in cohomology is defined as follows. For the cohomology-classrepresentative f ∈ HomH(F ′n, A

′′), there exists an f ∈ HomH(F ′n, A) which maps onto f via jH (bysurjectivity of jH). Then it is realized that ∂∗f = iH(f) for some f ∈ HomH(F ′n, A

′) which is unique

by injectivity of iH , and f is the class representative of δ[f ] where [f ] is the cohomology class of f .Thus in order for δ to commute with corestriction, it suffices to show that corestriction commutes with∂∗, i∗, j∗ where i∗ and j∗ refer to both maps associated to H,G. Let’s show commutativity of thediagram

HomG(Fn, A)jG // HomG(Fn, A

′′)

HomH(F ′n, A)jH //

cor

OO

HomH(F ′n, A′′)

cor

OO

Now cor[jH(f)] = cor[f = j f ] =∑g∈G/H gf(g−1 ) =

∑g∈G/H gj[f(g−1 )], and in the other di-

rection we have jG[cor(f)] = jG[∑g∈G/H gf(g−1 )] =

∑g∈G/H

gf(g−1 ) =∑g∈G/H j[gf(g−1 )] =∑

g∈G/H gj[f(g−1 )] where we note that j is a G-module homomorphism and hence commutes with theG-action. Thus the diagram is commutative, and similar calculations give commutativity with i∗ and ∂∗

since both are G-module homomorphisms.

42: Let p be a prime and let Sp be the symmetric group of degree p. Then each Sylow p-subgroupP is cyclic of order p, one such being the subgroup generated by the cycle (1 2 · · · p). Thus, H∗(P,Z) ∼=Z[ν]/(pν) where degν = 2. Show that NSp(P )/P ∼= Z∗p and that it acts on P ∼= Zp in the obvious way.Conclude that H∗(Sp,Z)(p)

∼= Z[νp−1]/(pν).

70

Since P is abelian, P ⊆ CSp(P ). Now |CSp(P )| = p(p − p)! = p as explained on pg127[2] wherewe note that P and its generator have the same centralizer, so we must have CSp(P ) ∼= P . Corol-lary 4.4.15[2] states that NSp(P )/CSp(P ) is isomorphic to a subgroup of Aut(P ), and by Proposition4.4.16[2] we know that Aut(P ) ∼= Z∗p is cyclic of order ϕ(p) = p − 1 [ϕ is the Euler function]. ThusNSp(P )/P ∼= H ⊆ Z∗p. The number of p-cycles in Sp is (p − 1)! as explained on pg127[2], and everyconjugate of P contains exactly p − 1 p-cycles, so there are (p − 1)!/(p − 1) = (p − 2)! conjugates of Pwhich is equal to the index |Sp : NSp(P )| by Proposition 4.3.6[2]; thus |NSp(P )| = p!/(p− 2)! = p(p− 1).This means |H| = |NSp(P )/P | = p − 1 and hence H ∼= Z∗p ⇒ NSp(P )/P ∼= Z∗p. This group actsby conjugation on P because it is a quotient of the normalizer, and this action is well-defined becauseconjugation by elements of P is trivial, noting that P is abelian.Theorem III.10.3[1] along with a theorem of Swan (see Exercise III.10.1) states that H∗(Sp,Z)(p)

∼=H∗(P,Z)NSp (P ). By Proposition II.6.2[1] the conjugation action by P induces the identity on H∗(P,Z), sothe resulting action (Corollary II.6.3[1]) is theNSp(P )/P -action induced onH∗(P,Z). ThusH∗(Sp,Z)(p)

∼=(Z[ν]/(pν))Z

∗p . For a particular dimension j, the elements of the jth-cohomology group belong to Zp and

hence the action on an element ν would be ν 7→ z∗ · ν = zν with 0 < z < p and z∗ ∈ Z∗p. Then in the

cohomology ring, the action is given by νi 7→ z∗ · νi = (z∗ · ν) ` · · · ` (z∗ · ν) = (zν)i = ziνi. So foran element to belong to the group of Z∗p-invariants, we must have νi = ziνi ⇒ zi ≡ 1 mod p for allz < p. This is only satisfied when i = p − 1 by Fermat’s Little Theorem (z is relatively prime to p), sothe invariant elements are of the form

∑i ziν

i(p−1). Therefore, H∗(Sp,Z)(p)∼= Z[νp−1]/(pν).

43: Prove that the Pontryagin product on H∗(G,Z) for G finite cyclic is the trivial map in positivedimensions (Z has trivial G-action).

The map is given by HiG ⊗ HjG → Hi+jG for each i, j. Since HnG ∼= G for n odd and HnG = 0for n even, the domain of the map [hence the map] is trivial for i or j even. But if both i := 2c + 1and j := 2c′+1 are odd, then i+j = 2(c+c′+1) is even, so the image of the map [hence the map] is trivial.

44: Compute H2r(Zp ⊕ Zq) where p and q are not necessarily relatively prime.

The cohomology Kunneth formula of Exercise V.2.2 gives H2r(Zp⊕Zq) ∼=[⊕2r

i=0Hi(Zp)⊗H2r−i(Zq)

]⊕[⊕2r+1

i=0 TorZ1(Hi(Zp), H2r−i+1(Zq)

)] ∼= [(Z⊗Zq)⊕ (Zp⊗Zq)r−1⊕ (Zp⊗Z)]⊕ [0] ∼= Zp⊕Zq ⊕Zr−1

gcd(p,q)

for r ≥ 1.The Tor-parts were trivial because each summand became Tor(Heven, Hodd) = 0 and Tor(Hodd, Heven) =0. Note that if p and q are relatively prime, then the result becomes Zp ⊕ Zq which agrees with the factthat Zp ⊕ Zq ∼= Zpq is cyclic.

45: Prove that Hi(Zn,Z) ∼= Z(ni

)for i, n ∈ N, where we interpret

(n<ii

)= 0. Deduce that Hi(Zn, k) ∼=

k

(ni

)for any commutative ring k.

For n = 1 we have H0(Z,Z) = H1(Z,Z) ∼= Z and Hi(Z,Z) = 0 for all i > 1 as proven on pg58[1].This agrees with the proposed formula, so we argue by induction on n (assume the result holds up to and

including n). The Kunneth formula of Exercise V.2.2 gives Hi(Zn+1) = Hi(Zn ⊕Z) ∼= [⊕i

p=0Hp(Zn)⊗

Hi−p(Z)] ⊕ [⊕i+1

p=0 TorZ1(Hp(Zn), Hi−p+1(Z)

)] ∼=

⊕ip=0H

p(Zn) ⊗ Hi−p(Z), where the latter isomor-

phism follows from the fact that TorZ1 (−,Z) = 0 and Hi−p+1(Z) is either Z or 0 as stated above. Now⊕ip=0H

p(Zn)⊗Hi−p(Z) ∼= (Hi(Zn)⊗Z)⊕ (Hi−1(Zn)⊗Z) ∼= Hi(Zn)⊕Hi−1(Zn) ∼= Z(ni

)⊕Z

(ni−1

), so

Hi(Zn+1,Z) ∼= Z(n+1i

)since

(ni

)+(ni−1

)= n!

i!(n−i)! + n!(i−1)!(n−i+1)! = n!(n−i+1)

i!(n−i+1)! + n!ii!(n−i+1)! = (n+1)!

i!(n+1−k)! =(n+1i

), and the inductive process is complete.

The cohomological analog of the universal coefficient sequence of Exercise III.1.3 gives Hi(Zn, k) ∼=(Hi(Zn)⊗ k)⊕ TorZ1 (Hi+1(Zn), k) ∼= (Z

(ni

)⊗Z k)⊕ 0 ∼= k

(ni

), where we note that TorZ1 (Zr,−) = 0.

Note: These integral cohomology groups are dual to the associated integral homology groups; see pg38[1].This observation of Poincare duality is reflected in the relation

(ni

)=(nn−i).

71

46: Let p be a prime, G a finite p-group, k a field of characteristic p, and n ∈ N. If Hn(G, k) = 0, provethat Hn+1(G,Z) = 0.

The cohomological analog of the universal coefficient sequence of Exercise III.1.3 gives Hn(G, k) ∼=(Hn(G) ⊗ k) ⊕ TorZ1 (Hn+1(G), k) = 0, which in particular implies TorZ1 (Hn+1(G), k) = 0. Then sincek contains the subring Zp, we must have TorZ1 (Hn+1(G),Zp) = 0 and hence no element in Hn+1(G)has order p. But |G| = pα and so pαHn+1(G) = 0 by Corollary III.10.2[1], which means any elementmust have order a power of p. The only way this is satisfied with no element having order p is for everyelement to have order 1 (if an element h had order pa with a ≥ 2, then pa−1h would be an element oforder p). Only the identity has order 1, so all elements are the same; thus Hn+1(G,Z) = 0.

47: Prove that the shuffle product is strictly anti-commutative.

The shuffle product on the bar resolution is given by[g1| · · · |gn] · [g1| · · · |gn] =

∑σ(−1)sgnσ[gσ−1(1)| · · · |gσ−1(n)|gσ−1(n+1)| · · · |gσ−1(2n)],

where σ ranges over the (n, n)-shuffles. But given any shuffled tuple (−1)s[gi1 | · · · |g| · · · |g| · · · |gi2n ] wehave a corresponding shuffle which simply swaps the two g’s, leaving the 2n-tuple fixed and altering thesign to (−1)s+1; the sign change arises from moving the left g an l amount of times and then moving theright g an l + 1 amount of times (the extra +1 is due to moving the right g around the left g) and thisgives a total of 2l + 1 amount of moves with (−1)s+2l+1 = (−1)s+1. So if there are an odd number ofshuffled tuples (due to n being even, i.e. the tuple is of even degree) then the sum will consist of pairedtuples with different signs plus one extra tuple, giving a nontrivial sum. But if the degree is odd (i.e.an even number of shuffled tuples) then the sum will consist of only paired tuples with different signs,giving a sum of 0. Thus x2 = 0 if degx is odd, where x = [g1| · · · |gn], and this is precisely the definitionof strict anti-commutativity.

48: Let p be a prime and let Cp be the cyclic group of order p with trivial Fp-action. Explain howthe fact H2(Cp,Fp) ∼= Fp and the classification of extensions of Cp by Fp matches up with the classifica-tion theorem for groups of order p2.

If P is a group with |P | = p2 then it has nontrivial center, |Z(P )| 6= 1, by Theorem 4.3.8[2]. If |Z(P )| = p2

then P = Z(P ) and P is abelian. The only other scenario is |Z(P )| = p, in which case |P/Z(P )| = p; thisimplies P/Z(P ) is cyclic and hence it is a fact that P is abelian. By the Fundamental Theorem of FinitelyGenerated Abelian Groups, P must then be either the cyclic group Cp2 or the elementary abelian groupCp × Cp; another proof of this is given in Corollary 4.3.9[2]. This means there are only two extensiongroups of Cp by Fp, but this does not necessarily mean that there are only two classes of extensions (pos-sible short exact sequences). Theorem IV.3.12[1] states that E(Cp,Fp) ∼= Fp and hence there are a totalof p classes of group extensions. There is the canonical split extension 0 → Fp → Cp × Cp → Cp → 1,and so the other extensions must fit into p − 1 classes and arise from projections Cp2 Cp (this isbecause the only injection Cp → Cp2 is the canonical inclusion, i.e. the pth-power map). Switching to

multiplicative notation, these extensions are 1 → Cp ≡ 〈a〉 → Cp2 ≡ 〈b〉βi→ Cp → 1 with a 7→ bp and

βi(b) = ai for 1 ≤ i ≤ p− 1.

49: Let G be a finite group, let H ⊆ G, and let K be any group. Let F be a field which acts triv-

ially on G and K, and consider the Kunneth isomorphism κ : H∗(G,F) ⊗F H∗(K,F)

∼=→ H∗(G ×K,F).Show that resG×KH×K κ = κ (resGH ⊗ id) and trG×KH×K κ = κ (trGH ⊗ id).

Both equations are straightforward, so we will only prove the latter (concerning the transfer map). ForuH ∈ H∗(H,F) and v ∈ H∗(K,F), κ is defined on H∗(H,F)⊗FH

∗(K,F) by κ(uH⊗v) = 〈uH×v, x⊗y〉 =〈uH , x〉 · 〈v, y〉, where we hide the factor (−1)degv·degx for convenience. Then (tr k)(uH ⊗ v) =tr〈uH × v, x ⊗ y〉 =

∑(g,k)∈(G×K)/(H×K)〈(g, k) · uH × v, (g, k)−1x ⊗ y〉 =

∑(g,1)∈(G/H)×1〈guH ×

1v, g−1x⊗1y〉 =∑g∈G/H〈guH , g−1x〉·〈v, y〉. But [κ(tr⊗id)](uH⊗v) = κ(

∑g∈G/H guH(g−1x)⊗v(y)) =∑

g∈G/H〈guH , g−1x〉 · 〈v, y〉, so the two compositions are in fact equal, as desired.

50: For an abelian group G whose order is divisible by the prime p, Theorem V.6.6[1] states that

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the isomorphism ρ :∧

Zp(Gp)⊗Zp ΓZp(pG)∼=→ H∗(G,Zp) is natural if p 6= 2. Why?

Referring to pg126[1], where pG = Tor(G,Zp) and Gp = G ⊗ Zp = G/pG, we have a split-exact uni-

versal coefficient sequence 0 →∧2

(Gp) → H2(G,Zp) → pG → 0. Now ρ(x ⊗ y) = ψ(x)ϕ(y), whereψ :

∧(Gp) → H∗(G,Zp) is the natural map of Theorem V.6.4[1] and ϕ : Γ(pG) → H∗(G,Zp) is the

Zp-algebra homomorphism extended from a splitting φ : pG→ H2(G,Zp) of the above sequence. Since ψis natural and ϕ is an extension of the splitting φ, the question of naturality of ρ reduces to the questionof naturality of φ (in dimension 2). The splitting is made by choice, and if p is odd, we may use the

canonical splitting H2(G,Zp) →∧2

(Gp) given in Exercise V.6.4(b) since 2 is invertible in Zp for p 6= 2(i.e. prime p odd); remember that a splitting can be made on either side of the sequence. Thus theisomorphism ρ is natural if p 6= 2. But if p = 2 then we do not have a known canonical splitting, and wecannot prove naturality in this case since we do not have an explicit map.

51: Referring to the proof of Proposition VI.2.6[1], if η : M → Q0 is an admissible injection (i.e. asplit injection of H-modules) and M is projective as a ZH-module, then why is Cokerη projective as aZH-module?

Since η is H-split, we have a split-exact sequence 0→Mη→ Q0 → Cokerη → 0, and so Q0 ∼= M⊕Cokerη.

But Q0 is ZG-projective by Corollary VI.2.2.[1], hence ZH-projective by Exercise I.8.2. Now a directsum is projective iff each direct summand is projective, so Cokerη must be ZH-projective.

52: Let G be a goup with order r. Show that for each q there exists G-modules C with Hq(G,C)cyclic of order r.

For q = 0 we can take C = Z since H0(G,Z) = Z/|G|Z = Zr. Then using the dimension-shifting tech-

nique, we can find G-modules C1 and C2 such that H0(G,Z) ∼= H1(G,C1) and H0(G,Z) ∼= H−1(G,C2);see property 5.4 on pg136[1]. Therefore, through repeatable applications of dimension-shifting, we can

range over all q to obtain Hq(G,C) ∼= Zr.

53: Fill in the details to the proof of Proposition VI.7.1[1] which states that the evaluation pairingρ : Hi(G,M ′)⊗Hi(G,M)→ Q/Z is a duality pairing, where M ′ = Hom(M,Q/Z).

Let F be a projective resolution of Z over ZG. The evaluation pairing is obtained by composing thepairing 〈·, ·〉 : HomG(F,M ′) ⊗ (F ⊗G M) → M ′ ⊗G M with the evaluation map M ′ ⊗G M → Q/Z.The pairing 〈u, z = x ⊗ m〉 is given by u ⊗ (x ⊗ m) = u(x) ⊗ m, and composing this with the evalu-ation map gives ρ(u ⊗ z) = [u(x)](m). Note that ρ gives rise to the map ρ : Hi(G,M ′) → Hi(G,M)′

defined by [ρ(u)](z) = ρ(u ⊗ z) = [u(x)](m). We have HomG(F,M ′) = HomG(F,Hom(M,Q/Z)) ∼=HomG(F⊗M,Q/Z)

(∗)= Hom(F⊗GM,Q/Z) = (F⊗GM)′. The equality (∗) arises from Exercise III.1.3, as

Q/Z has trivial G-action. Since ( )′ is exact, we can pass to homology to obtain H∗(G,M ′) ∼= H∗(G,M)′.This isomorphism resulted fromHomG(F,Hom(M,Q/Z)) ∼= HomG(F⊗M,Q/Z) which is given by Theo-rem 10.5.43[2], and this is precisely the map ρ. Therefore, since ρ is an isomorphism, ρ is a duality pairing.

54: Let G = Z2 = 〈g〉 and let A = Z8, written additively. Make G act on A by x 7→ 3x, and letG act trivially on B = Z2. Show that A is cohomologically trivial, but A⊗B is not.

First note that H∗(1,M) = 0 for any M ; it is clearly trivial in all dimensions not equal to −1 and 0,and in those two dimensions it is the kernel and cokernel of the norm map M →M which is the identity(so the kernel and cokernel are trivial). Thus for A to be cohomologically trivial it suffices to show that

H∗(G,A) = 0. The complete resolution from Exercise VI.3.1 implies that Hn(G,A) = CokerN for n even

and Hn(G,A) = KerN for n odd (see pg58[1]). The action on A gives AG = Z8/Z4 = 0+Z4, 1+Z4 ∼= Z2

since 1 7→ 3 ≡ 1 mod 2, and AG = Z2 = 0, 4 since 4 7→ 12 ≡ 4 mod 8. Thus the norm map N : Z2 → Z2

is given by 1 7→ N1 = 1 · 1 + g · 1 = 1 + 3 = 4, which is the identity. Thus KerN = CokerN = 0and H∗(G,A) = 0. However, repeating the above with coefficient module A ⊗ B we see that the normmap is the trivial map, 1 ⊗ 1 7→ 1 ⊗ 1 + g · (1 ⊗ 1) = 1 ⊗ 1 + 3 ⊗ 1 = 4 ⊗ 1 = 2 ⊗ 2 = 0. Thus

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H2i+1(G,A) = KerN = Z2 ⊗ Z2∼= Z2 and A⊗B is not cohomologically trivial.

55: Let H / G and let M be a G-module. Then MH is naturally a G/H-module, and the pair(ρ : G → G/H,α : MH → M) is compatible in the sense that α[ρ(g) · m] = g · α(m). Thus wehave a homomorphism inf : Hn(G/H,MH)→ Hn(G,M) called the inflation map.Using this, show that for the semi-direct product G = H oK and module M with trivial G-action, thegroup Hn(K,M) is isomorphic to a subgroup Hn(G,M).

From the inclusion i : K → G and the surjection ρ : G→ G/H = K we can pass to cohomology to ob-tain the restriction res : Hn(G,M)→ Hn(K,M) and the inflation inf : Hn(K,M)→ Hn(G,M). Sinceρ i = idK , the composite inf res is also the identity. Thus inf is injective, so Hn(K,M) ⊆ Hn(G,M)up to isomorphism.

55: In Exercise AE.34 it was shown that a module is finitely generated Z-projective iff it is finitelygenerated Z-free. Weakening the hypothesis, show that Z-projective = Z-free.

Free modules are projective, so it suffices to show that Z-projective implies Z-free. If P is Z-projectivethen it is a submodule of a Z-free module. But any submodule of a Z-free module is free by TheoremI.7.3[5], so P is Z-free.

57: Prove that the symmetric group S3 has periodic cohomology, and find its period.

Since S3∼= Z3 o Z2 and |S3| = 3! = 6, its Sylow subgroups are Z2 and Z3 which are both cyclic.

Then by Theorem VI.9.5[1], S3 has periodic cohomology.Alternatively, any proper subgroup must have order 1 or 2 or 3 and hence must be cyclic, so TheoremVI.9.5[1] implies that S3 has periodic cohomology.Alternatively, H4(S3,Z) ∼= Z6 by Exercise III.10.1 and so S3 has periodic cohomology by TheoremVI.9.1[1].We can see this periodicity via Exercise III.10.1, because Hn(S3) ∼= Hn+4(S3) for all n > 0. Thus theperiod is 4.

58: Let N : Z → ZG denote the G-module homomorphism z 7→ Nz, where N ∈ ZG is the normelement, and let ε : ZG → Z be the augmentation map. Prove that if α : Z → ZG is a G-modulehomomorphism then α = aN for some a ∈ Z, and if β : ZG → Z is a G-module homomorphism thenβ = bε for some b ∈ Z.

First note that α is determined by where it sends the identity, α(1) = x. Then for α to be com-patible with the G-action we must have g · x = g · α(1) = α(g · 1) = α(1) = x, so x ∈ (ZG)G = Z · Nwhere the equality is shown in Exercise AE.1; thus α = aN for some a ∈ Z. Now β is also determinedby β(1) = z, so we must have z = g · z = g · β(1) = β(g) = z′ and hence β maps G onto a single integerb; thus β = bε for some b ∈ Z (where we note that ε(G) = 1).

59: If G1 and G2 are perfect groups with universal central extensions E1 and E2, respectively, provethat E1 × E2 is a universal central extension of G1 ×G2.

We have universal central extensions 0 → H2(Gi) → Ei → Gi → 1 by hypothesis (see ExerciseIV.3.7). Since the direct sum of two extensions is an extension, we have a central extension 0 →H2(G1)⊕H2(G2)→ E1×E2 → G1×G2 → 1. I claim that this extension is a universal central extension.Indeed, H2(G1)⊕H2(G2) is isomorphic to H2(G1×G2) by the Kunneth formula, since H1(Gi) = 0 by per-fectness of Gi. Now, the universal central extension of G1×G2 is 0→ H2(G1×G2)→ E → G1×G2 → 1by definition, so E1 ×E2

∼= E by the Five-Lemma (applied to the two sequences of G1 ×G2) and henceE1 × E2 is the universal central extension of G1 ×G2.

60: In the proof of Proposition VIII.2.4[1], with Γ′ ⊂ Γ, where did we use the hypothesis that|Γ : Γ′| <∞?

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We considered a free Γ-module F , and noted that if F ′ is a free Γ′-module of the same rank thenF ∼= IndΓ

Γ′F′. We then applied Shapiro’s lemma to yield Hn(Γ′, F ′) ∼= Hn(Γ, F ). But Shapiro’s

Lemma is actually given by Hn(Γ′, F ′) ∼= Hn(Γ,CoindΓΓ′F

′). We therefore used the isomorphismCoindΓ

Γ′F′ ∼= IndΓ

Γ′F′ of Proposition III.5.9[1], which holds if |Γ : Γ′| <∞.

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11 References

[1] Kenneth Brown, Cohomology of Groups, Springer GTM 87 (1982).

[2] David Dummit and Richard Foote, Abstract Algebra (3rd Ed.), John Wiley & Sons (2004).

[3] Allen Hatcher, Algebraic Topology, Cambridge University Press (2006).

[4] James Munkres, Elements of Algebraic Topology, Perseus Books Publishing (1984).

[5] Serge Lang, Algebra (Revised 3rd Ed.), Springer GTM 211 (2002).

[6] James Munkres, Topology (2nd Ed.), Prentice Hall (2000).

Other Useful LiteratureAlejandro Adem & James Milgram, Cohomology of Finite Groups (2nd Ed.), Springer (2004).David Benson, Representations and Cohomology I+II, Cambridge University Press (1995).Jon Carlson & others, Cohomology Rings of Finite Groups (with Appendix), Springer (2003).Henri Cartan & Samuel Eilenberg, Homological Algebra, Princeton University Press (1956).Leonard Evens, The Cohomology of Groups, Oxford University Press (1991).Ross Geoghegan, Topological Methods in Group Theory, Springer GTM 243 (2008).Karl Gruenberg, Cohomological Topics in Group Theory, Springer (1970).Serge Lang, Topics in Cohomology of Groups, Springer (1996).Saunders Mac Lane, Homology, Springer (1995).Jean-Pierre Serre, Local Fields, Springer GTM 67 (1979).Jean-Pierre Serre, Galois Cohomology, Springer (2002).Charles Thomas, Characteristic Classes and Cohomology of Finite Groups, Cambridge University Press (1986).Edwin Weiss, Cohomology of Groups, Academic Press (1969).

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Solutions to Exercises from Kenneth Brown’s …cgerig/solns.pdfSolutions to Exercises from Kenneth Brown’s Cohomology of Groups Christopher A. Gerig, Cornell University (College