CHAPTER 1

Groups

1.1. Definitions and Easy Facts

Definition 1.1. A group is a nonempty set G equipped with a binary opera-tion · satisfying the following axioms.

(i) (associativity) (a · b) · c = a · (b · c) for all a, b, c ∈ G;(ii) (existence of identity) there exists e ∈ G such that ea = ae = a for all

a ∈ G;(iii) (existence of inverse) for each a ∈ G there exists a−1 ∈ G such that

aa−1 = a−1a = e.If ab = ba for all a, b ∈ G, G is called abelian. (Note. It is more appropriate totreat abelian groups as Z-modules.)

Facts. Let G be a group.(i) The identity e and the inverse of a ∈ G are unique. (Let e′ be another

identity of G and b another inverse of a. Then e′ = e′e = e and b = be =b(aa−1) = (ba)a−1 = a−1.)

(ii) (Cancelation) Let a, b, c ∈ G. Then ab = ac ⇒ b = c; ba = ca ⇒ b = c;ab = e⇒ a = b−1.

(iii) If a1, . . . , an ∈ G, parentheses are not needed to define a1 · · · an becauseof the associativity. Moreover, (a1 · · · an)−1 = a−1

n · · · a−11 .

Examples. Abelian groups: (Z,+), (R,+), (C,+), (R×, ·), (R× = R \ {0}),(C×, ·), Zn (integers modulo n).

Automorphism groups (groups of bijections which preserve a certain structure):SX = the group of all permutations of a set X; GL(V ) = the group of all invertiblelinear transformations of a vector space V ; GL(n, F ) = the group of n×n invertiblematrices over a field F , GL(n, F ) ∼= GL(Fn); the group of isometries of a metricspace; groups of automorphisms of groups, rings, modules, fields, graphs, etc.

Groups as invariants: the fundamental group and homology groups of a topo-logical space.

The law of exponents. Let G be a group, a ∈ G, m,n ∈ Z. Then aman =am+n, (am)n = amn. (Note. a0 := e; a−n := (a−1)n for n ∈ Z+.)

Additive notation (usually for abelian groups). We write a+ b for ab, 0 fore, −a for a−1 and na for an.

Cartesian product. If G and H are groups, the G × H is a group withoperation defined component wise.

Homomorphism. Let G and H be groups. A map f : G → H is called ahomomorphism if f(ab) = f(a)f(b) for all a, b ∈ G. It follows that f(eG) = eH and

1

2 1. GROUPS

f(a−1) = f(a)−1 for a ∈ G. ker f := f−1(eH) = {a ∈ G : f(a) = eH}. f is 1-1⇔ ker f = {eG}. If f : G → H is a homomorphism and a bijection, f is called anisomorphism. It follows that f−1 : H → G is also an isomorphism. If there existsan isomorphism f : G→ H, G and H are called isomorphic (G ∼= H). Isomorphicgroups have the same structure.

Aut(G) := the set of all automorphisms of G (isomorphisms from G to G).

(Aut(G), ◦) is the automorphism group of G.

Example. det : GL(n,R)→ R×, A 7→ detA, is a homomorphism.

Example. G = the set of all fractional linear transformations of C (functions

of the form f(z) =az + b

cz + d, where ad− bc 6= 0. (G, ◦) is a group.

φ : GL(2,C) −→ G[ a b

c d

]7−→ f,

where f(z) =az + b

cz + d, is a homomorphism with kerφ = {aI2 : a ∈ C×}.

Subgroups. Let (G, ·) be a group. H ⊂ G is called a subgroup of G, denotedH < G, if (H, ·) is a group. If X ⊂ G,

〈X〉 :=⋂

X⊂H<GH

is the smallest subgroup of G containing X; it is called the subgroup of G generatedby X.

〈X〉 = {ae11 · · · aenn : n ≥ 0, a1, . . . , an ∈ G, e1, . . . , en ∈ Z}.

If G = 〈a〉, G is called cyclic.

Example. Let F be a field and X ⊂ GL(n, F ) the set of all elementarymatrices. Then 〈X〉 = GL(n, F ).

Proposition 1.2. Let G be a group and ∅ 6= H ⊂ G.(i) H < G⇔ ab−1 ∈ H for all a, b ∈ H.(ii) If |H| <∞, then H < G⇔ H is closed under multiplication.

Proof. (ii) (⇐) Only have to show that H contains e and is closed underinversion. Let a ∈ H. The map x 7→ ax from H to H is 1-1 hence is onto. So,there exists x ∈ H such that ax = x. Thus, e = x ∈ H. Also, there exists y ∈ Hsuch that ay = e, so a−1 = y ∈ H. �

Cosets. Let H < G be groups and a ∈ G. aH = {ah : h ∈ H} is calleda left coset of H in G with representative a. Ha is a right coset of H in G.G/H := {aH : a ∈ G}, H\\G := {Ha : a ∈ G}. aH = bH ⇔ b−1a ∈ H;Ha = Hb⇔ ab−1 ∈ H. Cosets in G/H (H\\G) form a partition of G

Proposition 1.3.(i) |G/H| = |H\\G|. (|G/H| is called the index of H in G, denoted by

[G : H].)(ii) |G| = |G/H| |H|.

1.2. NORMAL SUBGROUPS AND QUOTIENT GROUPS 3

Proof. (i) α : G/H → H\\G, aH 7→ (aH)−1 = Ha−1, is a bijection.(ii) f : G → G/H, a 7→ aH, is onto. For each aH ∈ G/H, f−1(aH) = aH;

hence |f−1(aH)| = |aH| = |H|. So, |G| = |G/H| |H|. �

Corollary 1.4 (Lagrange’s theorem). If H < G and |G| <∞, then |H|∣∣ |G|.

For a ∈ G, the order of a is

o(a) := |〈a〉| =

{min{n ∈ Z+ : an = e} if there is n ∈ Z+ such that an = e,

∞ otherwise.

Fact. Let |G| <∞ and a ∈ G. Then o(a)∣∣ |G| and a|G| = e.

Example (Euler’s theorem). For n ∈ Z+, φ(n) := |{x ∈ Z : 0 < x ≤n, (x, n) = 1}|. We have

xφ(n) ≡ 1 (mod n) for all x ∈ Z and n ∈ Z+ with (x, n) = 1.

Proof. Let G = {x ∈ Zn : (x, n) = 1}. (G, ·) is a group of order φ(n). So,xφ(n) = 1 in G. �

Fact. Let H,K < G be groups. Then |H||K| = |HK||H ∩K|. If H and Kare finite, |HK| = |H||K|/|H ∩K|.

Proof. Define f : H × K → HK, (h, k) 7→ hk. For hk ∈ HK, f−1(hk) ={(ha, a−1k) : a ∈ H ∩K}; hence |f−1(hk)| = |H ∩K|. So, |H ×K| = |HK||H ∩K|. �

1.2. Normal Subgroups and Quotient Groups

Definition 1.5. Let H < G be groups. H is called a normal subgroup of G,denoted H C G, if aH = Ha for all a ∈ G. If H C G,

(G/H)× (G/H) −→ G/H

(aH, bH) 7−→ abH

is a well defined binary operation on G/H which makes G/H a group (the quotientor factor group of G by H.)

The canonical homomorphism. Let H C G. Then

π : G −→ G/H

a 7−→ aH

is an onto homomorphism with kerπ = H.

Easy fact. Let f : G→ H be a homomorphism and K < H. Then f(G) < Hand f−1(K) < G. If K /H, f−1(K) C G.

Easy facts. Let G be a group.(i) H C G⇔ H = ker f for some homomorphism f : G→ K.(ii) Z(G) := {a ∈ G : ax = xa ∀x ∈ G} (the center of G) is a normal subgroup

of G. If H < Z(G), H C G.(iii) If [G : H] = 2, H C G.

Proof. (iii) Let a ∈ G. If a ∈ H, aH = H = Ha. If a /∈ H, aH = G\H = Ha

since G = H�∪ aH = H

�∪Ha. �

4 1. GROUPS

Easy facts.(i) If H < G and K < G, then HK < G⇔ HK = KH.(ii) If H < G and K C G, then HK < G.(iii) If H C G and K C G, then HK C G.

Theorem 1.6 (The correspondence theorem). Let N C G and let A = {K <G : N ⊂ K}, B = the set of all subgroups of G/N . Then

f : A −→ BK 7−→ K/N

is a bijection. Moreover, K/N C G/N ⇔ K C G.

Universal mapping property of the quotient group. Let N C G andf : G→ H a group homomorphism such that ker f ⊃ N . Then ∃! homomorphismf : G/N → H such that the following diagram commutes.

................................................................................................. .......................................................................................... ............................

........................................................

.........................................

f

πf

G H

G/N

Proof. Let f(aN) = f(a). �

Three isomorphism theorems.

First isomorphism theorem. Let f : G → H be a homomorphism. ThenG/ ker f ∼= f(G).

Proof. By the universal mapping property, f : G → H induces a homo-morphism f : G/ ker f → f(G). f is onto and ker f = {ker f}. Hence f is anisomorphism. �

Second isomorphism theorem. Let H < G and N C G. Then HN/N ∼=H/(N ∩H).

Proof. f : H → HN/N , h 7→ hN , is an onto homomorphism with ker f =N ∩H. �

Third isomorphism theorem. Let N C G, H C G and N ⊂ H. Then(G/N)

/(H/N) ∼= G/H.

Proof. f : G/N → G/H, aN 7→ aH, is a well-defined onto homomorphismwith ker f = H/N . �

Example. If G is a cyclic group, then G ∼= Z or Zn.

Proof. Let G = 〈a〉. Then f : Z→ G, n 7→ an, is an onto homomorphism. Ifker f = {0}, G ∼= Z. If ker f 6= {0}, let n be the smallest positive integer in ker f .Then ker f = nZ. So, G ∼= Z/nZ = Zn. �

Fact. If (m,n) = 1, then Zm × Zn ∼= Zmn.

M1 Fact. If H,K C G, HK = G and H ∩K = {e}, then G ∼= H ×K.

1.2. NORMAL SUBGROUPS AND QUOTIENT GROUPS 5

Proof. For all h ∈ H, k ∈ K, we have hkh−1k−1 ∈ H∩K; hence hkh−1k−1 =e, i.e., hk = kh. Define f : H ×K → HK, (h, k) 7→ hk. Then f is an isomorphism.

�

Conjugation. For a ∈ G,

φa : G −→ G

x 7−→ axa−1

is an automorphism of G (conjugation by a). A subgroup H < G is normal iffφa(H) = H for all a ∈ G. H is called a characteristic subgroup of G if f(H) = Hfor all f ∈ Aut(G). “Characteristic”⇒ “normal”; the converse is false, e.g., Z2×{0}is a normal but not characteristic subgroup of Z2 × Z2.

Inn(G) := {φa : a ∈ G} is a normal subgroup of Aut(G) and is called the innerautomorphism group of G. (In fact, if f ∈ Aut(G), fφaf−1 = φf(a).) Moreover,Inn(G) ∼= G/Z(G). (Proof. g : G → Inn(G), a 7→ φa, is an onto homomorphismwith ker g = Z(G).)

Example. Let G be a group with |G| > 2. Then |Aut(G)| > 1.

Proof. If G is nonabelian, |Inn(G)| = |G/Z(G)| > 1. So, assume that G isabelian. If a2 6= e for some a ∈ G, x 7→ x−1 is a nontrivial automorphism of G. Ifa2 = e for all a ∈ G, G is a vector space over Z2 of dimension > 1. In this case,Aut(G) = GL(G). Let E be a basis of G and choose ε1, ε2 ∈ E distinct. Then∃f ∈ GL(G) such that f(ε1) = ε2. �

The commutator subgroup and abelianization. An element of the forx−1y−1xy (x, y ∈ G) is called a commutator of G. G′ = 〈{x−1y−1xy : x, y ∈ G}〉 isthe commutator subgroup of G.

Proposition 1.7.(i) G′ is a characteristic subgroup of G and G/G′ is abelian.(ii) Let H < G. Then H C G with G/H abelian ⇔ H ⊃ G′.

Note. G′ is the smallest normal subgroup H of G such that G/H is abelian. G/G′

is called the abelianization of G.

Proof. (i) {x−1y−1xy : x, y ∈ G} is invariant under any automorphism of G.(ii) (⇒) ∀x, y ∈ G, sinceG/H is abelian, (xH)(yH) = (yH)(xH); so, x−1y−1xy ∈

H.(⇐) H/G′ C G/G′ ⇒ H C G. G/H ∼= (G/G′)/(H/G′) is abelian. �

Normal subgroups at top and bottom. Let H < G. If H ⊂ Z(G) orH ⊃ G′, then H C G.

If G is abelian, every subgroup of G is normal. The converse is false, as shownby the next example.

Example. (The quaternion group) Q8 = {±1,±i,±j,±k}. The multiplicationin Q8 is defined by rules i2 = j2 = k2 = −1, ij = k, jk = i, ki = j, ik = −j,kj = −i, ji = −k; see Figure 1.1. Z(Q8) = {±1}. Every subgroup of Q8 isnormal, yet Q8 is nonabelian. Q8 = 〈{i, j}〉. Let A =

[0 1−1 0

], B = [ 0 i

i 0 ], C =[i 00 −i

]∈ GL(2,C). Then G = {±I,±A,±B,±C} < GL(2,C) and Q8

∼= G. (Theisomorphism is given by i 7→ A, j 7→ B.)

6 1. GROUPS

................................................................................

................................................................................................................................................................

................................

................................

................................

i

jk

ij = kjk = iki = j

................................................................................

................................................................................................................................................................

................................

................................

................................

i

jk

ik = −ikj = −iji = −k

Figure 1.1. Multiplication in Q8

Advanced Reading

Hamiltonian groups. A nonabelian group G is called Hamiltonian if all itssubgroups are normal.

Theorem 1.8 (Dedekind, Baer). G is Hamiltonian ⇔

G ∼= Q8 ×A×B,

where A is an elementary abelian 2-group and B is an abelian group whose elementsall have odd order.

Lemma 1.9. Let G be a group and x, y ∈ G such that [x, y] commutes with xand y. Then

(i) [xi, yj ] = [x, y]ij, i, j ∈ Z;(ii) (xy)i = xiyi[y, x](

i2), i ∈ N.

Proof. (i) First assume i, j ≥ 0. We have [x, y]2 = (x−1y−1xy)x−1y−1xy =x−1(x−1y−1xy)y−1xy = [x2, y]. Induction shows that [x, y]i = [xi, y]. In the sameway, [xi, y]j = [xi, yj ].

If i < 0, note that [x−1, y] = [x, y]−1.(ii) Induction on i.

(xy)i+1 = xiyi[y, x](i2)xy = xiyixy[y, x](

i2) = xixyi[yi, x]y[y, x](

i2)

= xi+1yi+1[y, x]i+(i2) = xi+1yi+1[y, x](

i+12 ).

�

Proof of Theorem 1.8. (⇐) Let H < G. Then H = (H ∩ (Q8×A))× (H ∩B). It suffices to show that H1 := H ∩ (Q8 × A) C Q8 × A. If H1 contains anelement of order 4, then H1 ⊃ {h2 : h ∈ H1} = Z(Q) = G′. So, H C G. If H1

contains no element of order 4, then H1 ⊂ Z(G). Also, H1 C G.(⇒) 1◦ Let x, y ∈ G such that c := [x, y] 6= 1. Since 〈x〉, 〈y〉 C G⇔ c ∈ 〈x〉∩〈y〉.

So, xi = c = yj for some i, j ∈ Z+. Put Q = 〈x, y〉. Then Q′ = 〈c〉 ⊂ Z(Q). ByLemma 1.9 (i), ci = [xi, y] = [c, y] = 1. Hence o(x), o(y) <∞.

2◦ In 1◦, assume o(x) + o(y) is minimal. Let p be a prime factor of o(x). Theminimality of o(x) + o(y) implies 1 = [xp, y] = cp. So, o(c) = p. Also, o(x) cannothave prime factors different from p. So, o(x) (and o(y)) are powers of p. Writexαp

r

= c = yβps

, α, β, r, s ∈ Z+, p - α, β. Then o(x) = pr+1, o(y) = ps+1. Letα′, β′ ∈ Z such that α′α ≡ 1 ≡ β′β (mod p). Then

xβ′pr

= xβ′α′αpr

= cα′β′ = yα

′β′βps

= yα′ps

,

where cα′β′ = [xβ

′, yα

′]. Replacing x and y by xβ

′and yα

′, we may assume xp

r

=c = yp

s

.

1.3. THE SYMMETRIC GROUP 7

3◦ In 2◦, assume r ≥ s. Since xpr−s

y does not commute with x, by the mini-mality of o(x) + o(y), o(xp

r−s

y) ≥ o(y) = ps+1. By Lemma 1.9 (ii),

1 6= (xpr−s

y)ps

= xpr

yps

[y, xpr−s

](ps

2 ) = [y, x]pr−s(ps

2 ) = c−12p

r(ps−1).

So, p - 12pr(ps − 1)⇒ p = 2 and r = 1. So, o(x) = 4, x2 = y2, yxy−1 = x−1 ⇒ Q is

a homomorphic image of Q8. Q is nonabelian ⇒ Q = Q8.4◦ G = QC where C = CG(Q). (Let g ∈ G. Then gxg−1 = x±1 = x(−1)a

,gyg−1 = y±1 = y(−1)b

, a, b ∈ {0, 1}. Then yaxbg commutes with x and y, i.e.yaxbg ∈ C.

5◦ C has no element of order 4. (Assume g ∈ C with o(g) = 4. Then g /∈ Q, soo(gx) = 2 or 4. Since [gx, y] 6= 1, ygxy−1 = (gx)−1, i.e. g = g−1. Contradiction.)

6◦ C is abelian. (Otherwise, C would be a Hamiltonian group without elementof order 4, which contradicts 3◦.) For every g ∈ G, gx does not commute withy. By 1◦, o(gx) < ∞ ⇒ o(g) < ∞. So, C is torsion ⇒ C = A × B, where Ais an elementary abelian 2-group and every element of B has odd order. ThenG = QC ∼= QA×B. Since A is a vector space over Z2, A = (Q ∩A)×A′ for someA′ < A. It’s easy to see that QA = Q×A′. �

1.3. The Symmetric Group

SX = the permutation group of X. If |X| = n, (may assume X = {1, . . . , n}),write SX = Sn. |Sn| = n!.

Cayley’s theorem. Every group G is isomorphic to a subgroup of SG. If|G| = n, G is embedded in Sn.

Proof. Define

f : G → SG

a 7→ f(a)where

f(a) : G → G

x 7→ ax.

f is a homomorphism with ker f = {e}. �

Notation. σ ∈ Sn is denoted by(1 · · · n

σ(1) · · · σ(n)

).

Let i1, . . . , ik ∈ {1, . . . , n} be distinct. (i1, . . . , ik) ∈ Sn is the permutation whichmaps i1 to i2, i2 to i3, ..., ik−1 to ik, ik to i1 and leaves other elements unchanged;it is a called a k-cycle. A transposition is a 2-cycle. A 1-cycle is the identity of Sn.

M1 Facts.(i) Disjoint cycles commute.(ii) (i1, . . . , ik) = (i2, . . . , ik, i1) = · · · .(iii) (i1, . . . , ik)−1 = (ik, . . . , i1).(iv) σ · (i1, . . . , ik) · σ−1 = (σ(i1), . . . , σ(ik)).

Theorem 1.10. M1 Every σ ∈ Sn is a product of disjoint cycles. The cyclesare unique up to their order in the product.

8 1. GROUPS

Proof. Existence. Induction on n. In the sequence 1, σ(1), σ2(1), . . . , let i beth smallest positive integer such that σi(1) = 1. Then, τ := (1, σ(1), . . . , σi−1(1))−1σfixes 1, σ(1), . . . , σi−1(1); hence τ ∈ S{1,...,n}\{σ(1),...,σi−1(1)}. By the induction hy-pothesis, τ is a product of disjoint cycles involving {1, . . . , n} \ {σ(1), . . . , σi−1(1)}.So, σ = (1, σ(1), . . . , σi−1(1))τ is a product of disjoint cycles.

Uniqueness. Assume σ = (1, i2, . . . , ik)α = (1, j2, . . . , jl)β, where α (β) is aproduct of disjoint cycles involving {1, . . . , n}\{1, i1, . . . , ik} ({1, . . . , n}\{1, j1, . . . , jl}).Then for 2 ≤ s ≤ min{k, l}, is = σs(1) = js. Claim that k = l. (If k < l, then1 = σ(ik) = σ(jk) = jk+1, which is a contradiction.) So, α = β. By induction, thedisjoint cycles in α and β are the same. �

Easy fact. If the disjoint cycles in σ ∈ Sn are of lengths l1, . . . , lk, theno(σ) = lcm(l1, . . . , lk).

M1 Facts. Sn is generated by each of the following subsets.(i) {(1, 2), (1, 3), . . . , (1, n)}.(ii) {(1, 2), (2, 3), . . . , (n− 1, n)}.(iii) {(1, 2), (1, 2, . . . , n)}.

Proof. (i) (i1, . . . , ik) = (i1, i2)(i2, i3) · · · (ik−1, ik); (i, j) = (1, i)(1, j)(1, i).(ii) (1, j) = (j − 1, j)(1, j − 1)(j − 1, j), 2 ≤ j ≤ n.(iii) (i, i+ 1) = (1, 2, . . . , n)(i− 1, i)(1, 2, . . . , n)−1. �

The sign of a permutation. Let f(x1, . . . , xn) be a polynomial in x1, . . . , xnwith coefficients in Z and let σ ∈ Sn. Define σ(f) = f(xσ(1), . . . , xσ(n)).

σ( ∏

1≤i<j≤n

(xi − xj))

=∏

1≤i<j≤n

(xσ(i) − xσ(j)) = ±∏

1≤i<j≤n

(xi − xj).

(Note.{{σ(i), σ(j)} : 1 ≤ i < j ≤ m

}is the set of all 2-element subsets of

{1, . . . , n}.) Write

σ( ∏

1≤i<j≤n

(xi − xj))

= sign(σ)∏

1≤i<j≤n

(xi − xj),

where sign(σ) ∈ {±1}. Then sign : Sn → ({±1}, ·) is a homomorphism (onto whenn ≥ 2). sign(1, 2) = −1; hence sign(i, j) = −1 since (i, j) is conjugate to (1, 2).An := ker sign is the alternating group of degree n.

Facts.(i) Each σ ∈ Sn is a product of either an even number of transpositions (not

necessarily disjoint) or an odd number of transpositions, but not both; σis called even or odd accordingly. An is the set of even permutations.

(ii) |An| = 12n! for n ≥ 2.

(iii) sign(i1, . . . , ik) = (−1)k−1.

Proposition 1.11. An is generated by 3-cycles.

Proof. Only have to show that (i, j)(k, l) (i 6= j, k 6= l) is generated by 3-cycles. If {i, j} = {k, l}, then (i, j)(k, l) = id. If |{i, j} ∩ {k, l}| = 1, may assume{i, j} = {1, 2}, {k, l} = {1, 3}. Then (1, 2)(1, 3) = (3, 2, 1). If {i, j}∩{k, l} = ∅, then(i, j)(k, l) = ((i, j)(j, k))((j, k)(k, l)) where |{i, j}∩{j, k}| = 1 = |{j, k}∩{k, l}|. �

1.3. THE SYMMETRIC GROUP 9

The dihedral group Dn. Dn is the subgroup of Sn generated by α =(1, . . . , n) and β = (2, n)(3, n− 1) · · · (dn2 e, b

n2 c+ 2).

Facts about Dn.(i) o(α) = n, o(β) = 2, βαβ−1 = α−1.(ii) For n ≥ 3, |Dn| = 2n. Dn = {αiβj : 0 ≤ i ≤ n− 1, 0 ≤ j ≤ 1}.(iii)

Z(Dn) =

{{id, αn/2} if n is even,{id} if n is odd.

Proof. (i) βαβ−1 = β(. . . , n−1, n, 1, 2, 3, . . . )β−1 = (. . . , 3, 2, 1, n, n−1, . . . ) =α−1.

(ii) By (i), {αiβj : 0 ≤ i ≤ n − 1, 0 ≤ j ≤ 1} is closed under multiplication.So, Dn = {αiβj : 0 ≤ i ≤ n − 1, 0 ≤ j ≤ 1}. αiβj , 0 ≤ i ≤ n − 1, 0 ≤ j ≤ 1,are all distinct. (Assume αi1βj1 = αi2βj2 , 0 ≤ i1, i2 ≤ n− 1, 0 ≤ j1, j2 ≤ 1. Thenαi1−i2 = βj2−j1 , 1 = βj2−j1(1) = αi1−i2(1) ≡ 1 + i1 − i2 (mod n). So, i1 = i2;hence j1 = j2.) �

Let Xn be a regular n-gon in R2 with vertices labeled by 1, 2, . . . , n (Figure 1.2).A symmetry of X is a rigid motion in R3 which permutes the vertices of Xn.

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........1

2

3

4

7

6

5

Figure 1.2. Regular heptagon

Fact. Dn = the group of symmetry of Xn.

Proof. Let σ ∈ Sn be a symmetry of Xn with σ(1) = i. Then α−iσ is asymmetry of Xn which fixes 1. Thus α−iσ is either the identity or the reflection β,i.e. α−iσ = βj , j = 0 or 1. So, σ = αiβj ∈ Dn. �

Definition 1.12. A group G 6= {e} is called simple if the only normal sub-groups of G are {e} and G.

Fact. A finite abelian group G is simple ⇔ G ∼= Zp where p is a prime.

Simplicity of An (n ≥ 5). If n ≥ 5, An is simple.

Proof. 1◦ If n ≥ 5, all 3-cycles in An are conjugate in An.Let i1, . . . , i5 ∈ {1, . . . , n} be distinct (arbitrary). Let

α =

(1 2 3 4 5i1 i2 i3 i4 i5

)or

(1 2 3 4 5i1 i2 i3 i5 i4

)

10 1. GROUPS

such that α ∈ An. Then α(1, 2, 3)α−1 = (i1, i2, i3).2◦ Let {id} 6= N C An. To show N = An, it suffices to show that N contains

a 3-cycle. (By the normality of N and 1◦, N contains all 3-cycles. By Proposi-tion 1.11, N = An.) Let id 6= σ ∈ N , written as a product of disjoint cycles.

Case 1. σ has a cycle of length r ≥ 4, say σ = (1, . . . , r) · α. Let δ = (1, 2, 3) ∈An. Then N 3 σ−1(δσδ−1) = (r, . . . , 2, 1)(2, 3, 1, 4, . . . , r) = (1, 3, r).

Case 2. σ has two 3-cycles, say, σ = (1, 2, 3)(4, 5, 6) · α. Let δ = (1, 2, 4) ∈ An.Then, N 3 σ−1(δσδ−1) = (3, 2, 1)(6, 5, 4)(2, 4, 3)(1, 5, 6) = (1, 4, 2, 6, 3). We are incase 1.

Case 3. One cycle of σ has length 3; all others have length ≤ 2. Then σ2 is a3-cycle.

Case 4. All cycles of σ has length ≤ 2. Say, σ = (1, 2)(3, 4)α. Let δ =(1, 2, 3) ∈ An. Then N 3 σ(δσδ−1) = (1, 2)(3, 4)(2, 3)(1, 4) = (1, 3)(2, 4) := τ . Letε = (1, 3, 5). Then N 3 τ(ετε−1) = (1, 3)(2, 4)(3, 5)(2, 4) = (1, 3, 5). �

Note. A1, A2, A3 are simple. A4 is not simple. K = {id, (1, 2)(3, 4), (1, 3)(2, 4),(1, 4)(2, 3)} C A4.

1.4. Group Actions

SX and XS. Let X be a set and σ a permutation of X. When applying σ tox ∈ X, we can write σ to the left or right of x, i.e. σ(x) or (x)σ. In the first notation,the rule of composition is (τσ)(x) = τ(σ(x)); the symmetric group on X is denotedby SX . In the second notation, the rule of composition is (x)(τσ) = ((x)τ)σ; thesymmetric group on X is denoted by XS. (Note. σ 7→ σ (σ 7→ σ−1) is an antiisomorphism (isomorphism) between SX and XS.)

Definition 1.13. Let G be a group and X a set. A left (right) action of Gon X is a homomorphism φ : G → SX (XS). Usually, for g ∈ G, the image φ(g)is still denoted by g. So, for a left (right) action, gx = φ(g)(x) (xg = (x)φ(g)).Equivalently, a left action on G on X is a map

G×X −→ X

(g, x) 7−→ gx

such that ex = x and g1(g2x) = (g1g2)x for all x ∈ X and g1, g2 ∈ G. A rightaction of g on X is a map X ×G→ X with similar properties.

Example. Let V be a vector space over F and HomF (V, F ) the dual of V .

GL(V )× V −→ V

(f, x) 7−→ f(x)

is a left action on GL(V ) on V .

HomF (V, F )×GL(V ) −→ HomF (V, F )(l, f) 7−→ l ◦ f

is a right action of GL(V ) on HomF (V, F ).

Definition 1.14. Let G act on X (left action). For x ∈ X, [x] := {gx : g ∈ G}is the G-orbit of x, Gx := {g ∈ G : gx = x} < G is the stabilizer of x. The G-orbitsform a partition of X. If X has only one G-orbit, we say that G acts transitivelyon X.

1.4. GROUP ACTIONS 11

M1 Easy facts. Let G act on X.(i) For x ∈ X, [G : Gx] = |[x]|.(ii) For g ∈ G and x ∈ X, Ggx = gGxg

−1.

Proof. (i) Define

f : G/Gx −→ [x]gGx 7−→ gx.

Then f is a well defined bijection. �

Conjugation and the class equation. G acts on G by conjugation: G×G → G, (a, x) 7→ axa−1. [x] is the conjugacy class of x in G. Gx = {a ∈ G :axa−1 = x} =: CG(x) is the centralizer of x in G. Note that |[x]| = 1 iff x ∈ Z(G).Let {xi : i ∈ I} be a set of representatives of the conjugacy classes of G notcontained in Z(G). Then [x], x ∈ Z(G), and [xi], i ∈ I, are all the conjugacyclasses of G. Hence

(1.1) |G| =∑

x∈Z(G)

|[x]|+n∑i∈I|[xi]| = |Z(G)|+

∑i∈I

[G : CG(xi)].

(1.1) is called the class equation of G.

Proposition 1.15. M1 Let p be a prime and |G| = pe > 1. (G is called afinite p-group.) Then Z(G) 6= {e}.

Proof. By the class equation, |Z(G)| ≡ 0 (mod p). �

M1 Fact. If |G| = p2 (p a prime), then G is abelian.

Proof. 1◦ Let G be any group. If G/Z(G) is cyclic, G is abelian.2◦ If |G| = p2, |Z(G)| = p or p2. However, |Z(G)| 6= p. (Otherwise, G/Z(G) is

cyclic ⇒ G is abelian ⇒ |Z(G)| = p2.) �

Normalizer. Let G be a group and S(G) the set of all subgroups of G. Gacts on S(G) by conjugation: G × S(G) → S(G), (a,H) 7→ aHa−1. For H < G,GH = {a ∈ G : aHa−1 = H} =: NG(H) is the normalizer of H in G. (NG(H)is the largest subgroup K of G such that H C K.) [G : NG(H)] is the number ofconjugates of H in G.

Proposition 1.16. Let G be a group. If there exists H < G such that [G :H] = n <∞, then there exists N C G such that N ⊂ H and [G : N ] | n!.

Proof. G acts on G/H: (a, gH) 7→ agH. Let N be the kernel of this action.Then N C G, N ⊂ H and G/N ↪→ SG/H = Sn. �

M1 Example. Let G be a simple group with |G| > 2. If ∃H < G such that[G : H] = n > 1, then G ↪→ An.

Proof. In the proof of Proposition 1.16, N = {e} and G = G/N ↪→ SG/N =Sn. Assume G ⊂ Sn. We claim that G ⊂ An. (If G 6⊂ An, then [G : G ∩ An] = 2.So, G ∩An C G,→←.) �

12 1. GROUPS

Theorem 1.17 (Burnside’s lemma). Let G be a finite group acting on a finiteset X. Then the number of G-orbits in X is

N =1|G|

∑g∈G

Fix(g).

where Fix(g) = |{x ∈ X : gx = x}|.

Proof. Let X = {(g, x) : g ∈ G, x ∈ X, gx = x}. Counting (g, x) ∈ X in theorder of g, x, we have |X | =

∑g∈G Fix(g). Counting (g, x) ∈ X in the order of x, g,

we also have |X | =∑x∈X |Gx|. Let [x1], . . . , [xn] be the G-orbits in X. Then∑

x∈X|Gx| =

n∑i=1

∑x∈[xi]

|Gx| =n∑i=1

∑x∈[xi]

|G||[x]|

=n∑i=1

∑x∈[xi]

|G||[xi]|

=n∑i=1

|G| = n|G|.

So, n|G| =∑g∈G Fix(g). Hence the theorem. �

1.5. Sylow’s Theorem

Theorem 1.18 (Sylow). M1 Let G be a finite group with |G| = pem, wherep is a prime, e > 0, and p - m.

(i) Every p-subgroup (including {e}) of G is contained in a subgroup of orderpe. (A subgroup of G of order pe is called a Sylow p-subgroup.)

(ii) All Sylow p-subgroups are conjugate in G.(iii) Let np denote the number of Sylow p-subgroups of G. Then np | m and

np ≡ 1 (mod p).

Proof. 1◦ Let X = {X ⊂ G : |X| = pe}. Then |X | =(pempe

)≡ m 6≡ 0

(mod p). G acts on X : G× X → X , (g,X) 7→ gX. There exists X ∈ X such that|[X]| 6≡ 0 (mod p), i.e. |G|/|GX | 6≡ 0 (mod p). So, pe

∣∣ |GX |. Choose x0 ∈ X.Then |GX | = |GXx0| ≤ |GXX| = |X| = pe. So, P := GX is a Sylow p-subgroup ofG.

2◦ Let P = {gPg−1 : g ∈ G}. P acts on P by conjugation. {P} is the onlyorbit with one element. (If P1 ∈ P such that P1 6= P and {P1} is an orbit, thenPP1 is a p-subgroup of G with |PP1| > pe, which is a contraction.) So, |P| ≡ 1(mod p). (Note: If a finite p-group acts on a set, the length of each orbit is a powerof p.) Since |P|

∣∣ |G|, we must have |P|∣∣ m.

3◦ Let P1 be a p-subgroup of G. P1 acts on P by conjugation. Since |P| ≡1 (mod p), there exists P2 ∈ P such that {P2} is a P1-orbit. Then P1 ⊂ P2.(Otherwise, P1P2 is a p-subgroup of G with |P1P2| > pe.) This also shows that Pis the set of all Sylow p-subgroups of G, i.e. np = |P|. �

Proposition 1.19. Let P be a Sylow p-subgroup of G. Then NG(NG(P )) =NG(P ).

Proof. Let x ∈ NG(NG(P )). Then xPx−1 ⊂ xNG(P )x−1 = NG(P ). Sinceboth xPx−1 and P are Sylow p-subgroups of NG(P ), by Sylow’s theorem, thereexists y ∈ NG(P ) such that xPx−1 = yPy−1 = P . So, x ∈ NG(P ). �

Proposition 1.20. Let H C G, |H| < ∞ and P a Sylow p-subgroup of H.Then G = H NG(P ).

1.5. SYLOW’S THEOREM 13

Proof. Let x ∈ G. Then P and xPx−1 are conjugate in H, i.e. xPx−1 =yPy−1 for some y ∈ H. So, y−1x ∈ NG(P ) and x = y(y−1x) ∈ HNG(P ). �

Note. The method in the above two proofs is called the Frattini argument.

Semidirect product. Let N and H be groups and θ : H → Aut(N) ahomomorphism. The semidirect product NoθH is a group where NoθH = N×Has a set and

(n1, h1)(n2, h2) =(n1[θ(h1)(n2)], h1h2

).

N ′ := N × {eH} C N oθ H, H ′ := {eG} × H < N oθ H; N ′ ∼= N , H ′ ∼= H;N oθ H = N ′H ′, N ′ ∩H ′ = {e}; H ∼= (N oθ H)/N ′.

Proposition 1.21. Let |G| = pq where p < q are primes.(i) If p - q − 1, G ∼= Zpq.(ii) If p | q−1, then G ∼= Zpq or G ∼= ZqoθZp, where θ : Zp → Aut(Zq) (∼= Z×q )

is any 1-1 homomorphism.

Proof. (ii) Assume G is nonabelian. Let Q be a Sylow q-subgroup and P aSylow p-subgroup of G. Then Q C G; hence G = QP . Write Q = 〈a〉, P = 〈b〉.Then bab−1 = ak, where k ∈ Z×q has multiplicative order p. Let [θ(1)](x) =lx∀x ∈ Zq. Since θ(1) ∈ Aut(Zq), l ∈ Z×q . Since θ : Zp → Aut(Zq) is 1-1,o(θ(1)) = o(1) (in Zp) = p. So, the multiplicative order of l in Z× is p. We use afact that Z× is cyclic (cf. Fact ??). Since k, l ∈ Z×q are both of order p, we can writel = ks, p - s. (Cf. Exercise 1.3 (iv).) Let c = bs. Then P = 〈c〉 and cac−1 = al.Then

φ : Zq oθ Zp −→ G

(i, j) 7−→ aicj

is an isomorphism. In fact,

φ((i1, j1)(i2, j2)

)= φ(i1 + [θ(j1)](i2), j1 + j2)

= φ(i1 + [θ(1)j1 ](i2), j1 + j2

)= φ(i1 + lj1i2, j1 + j2)

= ai1+lj1 i2cj1+j2

= ai1(cj1ac−j1)i2cj1+j2 (∵ cj1ac−j1 = alj1 )

= ai1bj1ai2c−j1cj1+j2

= ai1bj1ai2cj2

= φ(i1, j1)φ(i2, j2).

�

Example. Let |G| = pqr, where p < q < r are primes. Then the Sylowr-subgroup of G is normal.

Proof. Let Q < G, R < G such that |Q| = q, |R| = r. Assume to thecontrary that R 6C G. Then nr = pq, i.e. NG(R) = R. If Q C G, then R C QR,NG(R) ⊃ QR, which is a contradiction. If Q 6C G, nq = r or pr. Then G haspq(r − 1) elements of order q and at least r(q − 1) elements of order q. Note thatpq(r − 1) + r(q − 1) > pqr, contradiction. �

Example. If G is a simple group of order 60, then G ∼= A5.

14 1. GROUPS

Proof. 1◦ Let P be the set of all Sylow 5-subgroups of G. Then |P| = 6. LetG act on P by conjugation. Then G ↪→ SP = S6. Thus G ↪→ A6. (Otherwise,[G : G ∩A6] = 2⇒ G ∩A6 C G.) So, we assume G ⊂ A6.

2◦ We claim that ∃ θ ∈ Aut(A6) such that θ(G) = {σ ∈ A6 : σ(1) = 1} ∼= A5.Define

φ : A6 −→ SA6/G

σ 7−→ φ(σ),

whereφ(σ) : A6/G −→ A6/G

αG 7−→ σαG.

Then φ is a homomorphism with kerφ ⊂ G. Hence kerφ = {id} and φ : A6 ↪→SA6/G. Since A6 is simple, φ : A6

∼=→ AA6/G. Write A6/G = {G1, . . . , G6}, whereG1 = G. Let f : {1, . . . , 6} → A6/G, i 7→ Gi. Define

θ : An −→ An

σ 7−→ f−1φ(σ)f.

Then θ ∈ Aut(A6). If σ ∈ G, [θ(σ)](1) = [f−1φ(σ)f ](1) = [f−1φ(σ)](G) =f−1(G) = 1. �

1.6. Finitely Generated Abelian Groups

If A and B are abelian groups, A×B is also denoted by A⊕B.

Theorem 1.22 (Structure of finitely generated abelian groups). Every finitelygenerated abelian group G is isomorphic to

Zpe11⊕ · · ·Zpes

s⊕ Zr,

where p1, . . . , ps are primes (not necessarily distinct), ei > 0, r ≥ 0. The num-bers pe11 , . . . , p

ess (the elementary divisors of G) and r (the rank of G) are uniquely

determined by G.

Proof. This is a special case of Theorem 2.55, the structure theorem of finitelygenerated modules over a PID. �

Facts.

(i) The number of nonisomorphic abelian groups of order pe11 · · · pess is

P (e1) · · ·P (es), where P (e) is the number of partitions of e.(ii) Assume A,B,C are finitely generated abelian groups. Then A ⊕ A ∼=

B⊕B ⇒ A ∼= B; A⊕C ∼= B⊕C ⇒ A ∼= B. (Use the structure theorem.)These are false if A,B,C are not finitely generated. Let A = Zℵ0 andB = Z. Then A⊕A ∼= B⊕A but A 6∼= B. Corner’s theorem (next) showsthat there is an abelian group A such that A⊕A⊕A ∼= A but A⊕A 6∼= A.Let B = A⊕A. Then B ⊕B ∼= A⊕A but B 6∼= A.

Advanced Reading

Theorem 1.23 (Corner [3]). Let r ∈ Z+. There exists a countable reducedtorsion-free abelian group G such that Gm ∼= Gn ⇔ m ≡ n (mod r). (An abeliangroup is called reduced if its only divisible subgroup is {0}.)

1.6. FINITELY GENERATED ABELIAN GROUPS 15

Lemma 1.24 (Corner [2]). Let A be a ring such that (A,+) is countable, re-duced, and torsion-free. Then A ∼= End(G) for some countable, reduced, torsion-freeabelian group G.

Lemma 1.25. Let G be an abelian group and ω1, ω2 two idempotents in End(G).Then ω1G ∼= ω2G⇔ there exist x, y ∈ End(G) such that ω1 = xy and ω2 = yx.

Proof. (⇐) Let y∗ : ω1G → ω2G, ω1a 7→ yω1a and x∗ : ω2G → ω1G, ω2a 7→xω2a. Then x∗ and y∗ are inverses of each other.

(⇒) Let α : ω1G → ω2G and β : ω2G → ω1G be inverse isomorphisms. Lety = αω1 and x = βω2. �

Proof of Theorem ??. Only have to consider r > 1. Let A be the ringfreely generated by xi, yi, 0 ≤ i ≤ r, subject to the relation

yixj =

{1 if i = j,

0 if i 6= j,

r∑i=0

xiyi = 1.

1◦ (A,+) is free abelian of countable rank.We have A ∼= B/C, where B is the ring freely generated by xi, yi, 0 ≤ i ≤ r,

subject to the relations

yixj =

{1 if i = j,

0 if i 6= j

and C is the ideal of B generated by 1−∑ri=0 xiyi.

We claim that B is free abelian with a basis

(1.2) xi1 · · ·ximyjn · · · yj1 , n,m ≥ 0, 0 ≤ ik, jk ≤ r.Let A be the set of elements in (1.2), B the set of products of elements of

{x1, . . . , xr, y1, . . . , yr} containing at least one string yixj (i 6= j) but no stringyixi, and C the set of products of elements in {x1, . . . , xr, y1, . . . , yr} ∪ {yixi − 1 :1 ≤ i ≤ r} containing at least one yixi − 1 but no string yixi. Then A∪ B ∪ C is aZ-basis of the free ring generated by {x1, . . . , xr, y1, . . . , yr}. To see this, it sufficesto show that A ∪ B ∪ C is linearly independent over Z. Assume

(1.3) α1a1 + · · ·+ αsas + β1b1 + · · ·+ βtbt + γ1c1 + · · ·+ γucu = 0,

where ai ∈ A, bj ∈ B, ck ∈ C, αi, βj , γk ∈ Z. Assume that c1 has the largest numberof elements in {x1, . . . , xr, y1, . . . , yr} (counted with multiplicity). In (1.3), expandeach ck as a product of elements in {x1, . . . , xr, y1, . . . , yr}. Let c′1 be obtained fromc1 by replacing each yixi − 1 with yixi. Then in the LHS of (1.3), γ1c

′1 is the only

term of involving c′1. Therefore, γ1 = 0. In the same way, γ1 = · · · = γu = 0. Itfollows immediately that α1 = · · · = αs = β1 = · · · = βt = 0.

Since B is the free ring on {x1, . . . , xr, y1, . . . , yr} modulo the ideal generatedby B ∪ C, B is free abelian with a basis A.

C is generated by

xi1 · · ·xim(1−

r∑i=0

xiyi

)yjn · · · yj1 , n,m ≥ 0, 0 ≤ ik, jk ≤ r.

It can be seen that B = C ⊕D, where

D := 〈xi1 · · ·ximyjn · · · yj1 , n,m ≥ 0, 0 ≤ ik, jk ≤ r, (im, jn) 6= (r, r)〉.Thus B/C ∼= D.

16 1. GROUPS

2◦ By Lemma 1.24, A = End(G) for some countable, reduced, torsion-freeabelian group G. Let ωi = xiyi, 0 ≤ i ≤ r. Then 1 = ω0+· · ·+ωr is a decompositionof 1 into orthogonal idempotents. So,

G =r∑i=0

ωiG.

Since ωi = xiyi and 1 = yixi, by Lemma 1.25, G ∼= ωiG. Thus G ∼= Gr+1; henceGm ∼= Gn if m ≡ n (mod r).

3◦ There exists Z-linear map T : A→ Zr such that(i) T (xy) = T (yx), x, y ∈ A;(ii) T (ωi) = 1, 0 ≤ i ≤ r.

We only have to construct a Z-linear map T ∗ : B → Zr satisfying the analogiesof (i) and (ii) and T ∗|C = 0. Define

T ∗(xi1 · · ·ximyjn · · · yj1) =

{1 if m = n and ik = jk, 1 ≤ k ≤ m,0 otherwise

and extend T ∗ to B by linearity.4◦ Gm 6∼= Gn if m 6≡ n (mod r).Assume Gm ∼= Gn for some 1 ≤ m,n ≤ r. Then(m−1∑

i=0

ωi

)G =

m−1∑i=0

ωiG ∼=n−1∑i=0

ωiG =(n−1∑i=0

ωi

)G.

By Lemma 1.25, there exist x, y ∈ A such that∑m−1i=0 ωi = xy,

∑n−1i=0 ωi = yx.

Then in Zr,

m = T(m−1∑i=0

ωi

)= T (xy) = T (yx) = T

(n−1∑i=0

ωi

)= n.

�

1.7. Free Groups

FX , the free group on X. Let X be a set and

W ={xe11 · · ·xes

s : s ≥ 0, xi ∈ X, ei ∈ {±1}}.

(Here, xe11 · · ·xess is a formal product and is called a word in X.) For u, v ∈ W, say

u ∼ v if v can be obtained form u by deleting and inserting strings of the form xx−1

or x−1x, x ∈ X. ∼ is an equivalence relation on W; the equivalence class of u ∈ Wis denoted by [u]. Put FX = W/∼ and for [u], [v] ∈ W/∼, define [u][v] = [uv],where uv is the concatenation of u and v. This gives a well defined operation onFX ; FX with this operation is a group, called the free group on X; rankFX := |X|.

Let G be any group and f : X → G. Then there is a unique homomorphismf : FX → G such that

.................................................................................................................. ..........................................................................................

................................................................................................. ...........f

ιf

F

GX

1.7. FREE GROUPS 17

commutes, where ι(x) = [x], x ∈ X.

Every group is a homomorphic image of a free group.A word w ∈ W is called reduced if it contains no strings xx−1 or x−1x, x ∈ X.

Fact. Every class in W/∼ contains a unique reduced word.

Proof. (Uniqueness) Let R be the set of all reduced words in W. Define

f : X −→ SR

x 7−→ f(x)

wheref(x) : R −→ R

xe11 · · ·xess 7−→

{xxe11 · · ·xes

s if xe11 6= x−1,

xe22 · · ·xess if xe11 = x−1.

There exists homomorphism f : FX → SR such that f([x]) = f(x) for all x ∈ X.Note that for x ∈ X, we have

f(x)−1 : R −→ R

xe11 · · ·xess 7−→

{x−1xe11 · · ·xes

s if xe11 6= x,

xe22 · · ·xess if xe11 = x.

Assume xe11 · · ·xess , y

f11 · · · y

ft

t ∈ R such that [xe11 · · ·xess ] = [yf11 · · · y

ft

t ]. Thenxe11 · · ·xes

s =(f(x1)e1 · · · f(xs)es

)(1) = f([xe11 · · ·xes

s ])(1) = f([yf11 · · · yft

t ])(1) =(f(y1)f1 · · · f(yt)ft

)(1) = yf11 · · · y

ft

t . �

So, ι : X → FX is an injection and X can be regarded as a subset of FX .

Presentation. Let X be a set and R a set of words in X. Define

〈X | R〉 = FX/N,

where N is the normal closure of {[r] : r ∈ R} in FX , i.e., the smallest subgroup ofFX containing {[r] : r ∈ R}. If G ∼= 〈X | R〉, 〈X | R〉 is called a presentation of G;elements in X are called generators; elements in R are called relations.

Theorem 1.26 (Van Dyck). Let G be a group generated by X ⊂ G. Let R bea set of words in X such that for every r ∈ R, r = 1 in G, i.e., the relation r = 1is satisfied in G. Then ∃! onto homomorphism 〈X | R〉 → G such that [x] 7→ x forall x ∈ X.

Proof. Let N be the normal closure of {[r] : r ∈ R} in FX . The map f :X → G, x 7→ x, induces a homomorphism f : FX → G such that [x] 7→ x. SinceG = 〈X〉, f is onto. By assumption, {[r] : r ∈ R} ⊂ ker f . So, N ⊂ ker f . Thus, finduces an onto homomorphism from FX/N = 〈X | R〉 to G.

The uniqueness is obvious since 〈X | R〉 is generated by {[x]N, : x ∈ X}. �

Examples.• Dn = 〈α, β | αn = β2 = 1, βαβ−1 = α−1〉.• The infinite dihedral group

D∞ = 〈α, β | β2 = 1, βαβ−1 = α−1〉.

18 1. GROUPS

• The quaternion group

Q8 = 〈x, y | x4 = 1, y2 = x2, yxy−1 = x−1〉.

Proof. Q8 = 〈i, j〉, where i, j satisfies relations i4 = 1, j2 = i2, jij−1 =i−1. By Van Dyck’s theorem, ∃ onto homomorphism

(1.4) G := 〈x, y | x4 = 1, y2 = x2, yxy−1 = x−1〉 −→ Q8

such that x 7→ i, y 7→ j. Every element in G can be written as xiyj ,0 ≤ i ≤ 3, 0 ≤ j ≤ 1. So, |G| ≤ 8. Hence the homomorphism in (1.4) isan isomorphiam. �

• The generalized quaternion group

Q2n = 〈x, y | x2n−1= 1, y2 = x2n−2

, yxy−1 = x−1〉.

The braid group on n strings.

Bn = 〈σ1, . . . , σn−1 | σiσi+1σi = σi+1σiσi+1, 1 ≤ i ≤ n−2; σiσj = σjσi, |i−j| > 1〉.

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σ1σ2σ1 σ2σ1σ2=

=

Figure 1.3. The braid group

Fact. Let X = A�∪ B, H the subgroup of FX generated by A and N the

normal closure of B in FX . Then FX = HN and H ∩ N = {1}. In particular,F/N ∼= H is free on A.

Proof. Only have to show H ∩N = {1}. Define

φ : X −→ FA

x 7−→

{x if x ∈ A,1 if x ∈ B.

φ induces a homomorphism φ : FX → FA such that N ⊂ ker φ and φ|H : H → FAis an isomorphism. If w ∈ H ∩N , then φ|H(w) = φ(w) = 1, hence w = 1. �

Fact. FX1∼= FX2 ⇔ |X1| = |X2|.

Proof. (⇒) FXi/(FXi

)′ is free abelian of rank |Xi|. �

Theorem 1.27 (The Nielson-Schreier theorem). Let F be a free group on X andG < F . Then G is a free group. If [F : G] = m <∞, then rankG = m|X|+1−m.(|X| may be infinite.)

Advanced Reading

1.7. FREE GROUPS 19

Proof. (Based on Weir [11]; see Rotman [10] and Robinson [9] for moredetails.)

1◦ Let τ be a (right) transversal (function) of G in F , i.e., τ is a function fromG\\F to F such that π ◦ τ = idG\\F , where π : F → G\\F is the canonical map.Always assume that τ(G) = 1. Define

α : F × (G\\F ) −→ G

(u,C) 7−→ τ(C)uτ(Cu)−1.

Let F be the free group on X× (G\\F ). φ := α|X×(G\\F ) induces a homomorphismφ : F → G.

2◦ We construct a map ψ : F × (G\\F ) → F such that the following diagramcommutes.

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......................................................................................................................................................................................................................................................................................... ............................................................................................................ ...........α

φψ

φ=α|X×(G\\F )

G

F

X × (G\\F )

F × (G\\F )

Define

ψ : F × (G\\F ) −→ F

(u,C) 7−→ uC

where uC is defined inductively on the length of u:

uC =

1 if u = 1,(x,C) if u = x ∈ X,(x,Cx−1)−1 if u = x−1, x ∈ X,vCyCv if u = vy is reduced and y ∈ X ∪X−1.

Then for all u, v ∈ F and C ∈ G\\F , (uv)C = uCvCu, (u−1)C = (uCu−1

)−1 and

φ(uC) = τ(C)u τ(Cu)−1.

3◦ Letβ : G −→ F

u 7−→ uG

β is a homomorphism and φ ◦ β = idG. So, φ is onto, β is 1-1, and

G ∼= F / ker φ.

4◦ Let N be the normal closure of {τ(C)G : C ∈ G\\F} in F . We claim thatker φ = N .

φ(τ(C)G) = τ(G)τ(C)τ(C)−1 = 1. So, N ⊂ ker φ. To show that ker φ ⊂N , note that ker φ = ker γ, where γ = β ◦ φ. It suffices to show that γ(xC) ≡xC (mod N) for (x,C) ∈ X × (G\\F ). We have γ(xC) =

[τ(C)x τ(Cx)−1

]G =

τ(C)GxC(τ(Cx)−1

)Gτ(C)x = τ(C)GxC(τ(Cx)G

)−1 ≡ xC (mod N).

20 1. GROUPS

5◦ There exists a (right) transversal τ of G in F (τ(G) = 1) such that if areduced word uy ∈ im(τ), y ∈ X ∪X−1, then u ∈ im(τ). im(τ) is said to have theSchreier property and τ is called a Schreier transversal. (For each C ∈ G\\F , thelength of C, denoted by |C|, is defined to be the minimum length of words in C.Define τ(C) inductively on |C|. τ(G) = 1. Assume that τ(D) have been defined forall D ∈ G\\F with |D| < |C|. Let uy ∈ C (y ∈ X ∪X−1) be of minimum length.Define τ(C) = τ(Gu)y.)

6◦ Let τ be a Schreier transversal of G in F . For each G 6= C ∈ G\\F ,τ(C) = uxe (reduced), x ∈ X, e ∈ {±1}. So, τ(C)G = uG(xe)Gu, where

(xe)Gu =

{xGu if e = 1(xGux

−1)−1 if e = −1

}= (xD)e for some D ∈ G\\F.

Since im(τ) has the Schreier property, u ∈ im(τ), so, uG ∈ N hence xD ∈ N .Applying the same argument to u and use induction on the length of u, we see thatτ(C)G is a product of certain elements in N of the form xD (x ∈ X, D ∈ G\\F )and their inverses. So, N is the normal closure of N ∩

(X × (G\\F )

). Then

G ∼= F /N (by 3◦, 4◦)∼= the free group on X × (G\\F )

)\N.

7◦ Assume [F : G] = m <∞. To show that rankG = m|X|+ 1−m, it sufficesto show that ∣∣N ∩ (X × (G\\F )

)∣∣ = m− 1.

Note that for xC ∈ X × (G\\F ), x ∈ N (= ker φ)⇔ τ(C)x τ(Cx)−1 = 1. Define

θ : (G\\F ) \ {G} −→ N ∩(X × (G\\F )

)C 7−→ θ(C)

where

θ(C) =

{xGu if τ(C) = ux (reduced) x ∈ X,xC if τ(C) = ux−1 (reduced) x ∈ X.

It is easy to check that θ(C) ∈ N and θ is 1-1. (Needs the Schreier property.) GivenxC ∈ N (x ∈ X),

xC =

{θ(C) if τ(C) (reduced) ends with x−1,

θ(Cx) otherwise.

So, θ is also onto. �

1.8. Nonabelian Groups of Order ≤ 30

Order pq, p | q−1. Orders 6, 10, 14, 21, 22, 26 are covered by Proposition ??.

Order 8. G ∼= Q8 or D4.

Proof. G has no element of order 8. G has an element of order 4. (Otherwise,x2 = e ∀x ∈ G. Then G is abelian.) Let a ∈ G such that o(a) = 4. Chooseb ∈ G \ 〈a〉. Then G = 〈a, b〉. Since |G/〈a〉| = 2, b2 ∈ 〈a〉; hence b2 = a2 or e sinceo(b2) = 1 or 2. Since 〈a〉 C G, bab−1 ∈ 〈a〉; hence bab−1 = a−1.

Case 1. b2 = a2. Then G ∼= Q8.

1.8. NONABELIAN GROUPS OF ORDER ≤ 30 21

Table 1.1. Nonabelain groups of order ≤ 30

|G| G

6 S3

8 D4, Q8

10 D5

12 A4, D6, T = Z3 oα Z4

14 D7

16D8, Q16, D4 × Z2, Q8 × Z2, Z8 oβ1 Z2, Z8 oβ2 Z2,

(Z4 × Z2) oβ3 Z2, (Z4 × Z2) oβ4 Z2, Z4 oβ5 Z4

18 D9

20 D10, Z5 oγ1 Z4, Z5 oγ2 Z4

21 Z7 o Z3

22 D11

24D4 × Z3, Q8 × Z3, Z3 oδ1 Z8, Z3 oδ2 (Z4 × Z2),

Z3 oδ3 (Z4 × Z2), Z2 ×D6, Z3 oδ4 (Z2 × Z2 × Z2),Q8 oδ5 Z3, Z3 oδ6 Q8, Z3 oδ7 D4, Z3 oδ8 D4, S4

26 D13

27 Z9 o Z3, (Z3 × Z3) o Z3

28 D14, Z7 o Z4

30 D15, D5 × Z3, S3 × Z5

Case 2. b2 = e. Then G ∼= D4. �

Order 12. Let a = ((1, 2, 3), 2), b = ((1, 2), 1) ∈ S3×Z4. T := 〈a, b〉 < S3×Z4

is a subgroup of order 12.

T = 〈a, b | a6 = 1, a3 = b2, bab−1 = a−1〉 = {aibj : 0 ≤ i ≤ 5, 0 ≤ j ≤ 1}.

Let x = a2 and y = b, we also have

T = 〈x, y | x3 = y4 = 1, yxy−1 = x−1〉 ∼= Z3 oα Z4,

where α : Z4 → Aut(Z3), [α(1)](1) = −1.

Order 16. Talbe 1.2. (Cf. [1, §118].)

Order 20. G ∼= D10, Z5 oγ1 Z4, Z5 oγ2 Z4, where γ1, γ2 : Z4 → Aut(Z5),[γ1(1)](1) = 2, [γ2(1)](1) = −1.

Proof. Let 〈a〉 C G be the Sylow 5-subgroup and P a Sylow 2-subgroup. IfP = 〈b〉〈c〉, where o(b) = o(c) = 2, we may assume bab−1 = a. (If bab−1 = a−1

and cac−1 = a−1, then (bc)a(bc)−1 = a.) Then G = 〈ab〉〈c〉, where Z10∼= 〈ab〉 C

G. Thus G ∼= Z10 o Z2∼= D10. If P = 〈b〉 ∼= Z4, then G ∼= Z5 oγ Z4, where

γ : Z4 → Aut(Z5) is a homomorphism such that [γ(1)](1) 6= 1. If [γ(1)](1) = 2 or3, G ∼= Z5 oγ1 Z4. If [γ(1)](1) = −1, G ∼= Z5 oγ2 Z4.

Z5 oγ1 Z4 6∼= Z5 oγ2 Z4 since Z(Z5 oγ1 Z4) = {(0, 0)} but Z(Z5 oγ2 Z4) =〈(0, 2)〉 �

22 1. GROUPS

Table 1.2. Nonabelian groups of order 16

group presentation

D8 〈x, y | x8 = y2 = 1, yxy−1 = x−1〉Q16 〈x, y | x8 = 1, y2 = x4, yxy−1 = x−1〉

D4 × Z2〈x, y, z | x4 = y2 = z2 = 1, y−1xy = x−1,

xz = zx, yz = zy〉

Q8 × Z2〈x, y, z | x4 = z2 = 1, y2 = x2,

y−1xy = x−1, xz = zx, yz = zy〉Z8 oβ1 Z2,

β1 : Z2 → Aut(Z8), [β1(1)](1) = 5.〈x, y | x8 = y2 = 1, y−1xy = x5〉

Z8 oβ2 Z2

β1 : Z2 → Aut(Z8), [β1(1)](1) = −1.〈x, y | x8 = y2 = 1, y−1xy = x−1〉

(Z4 × Z2) oβ3 Z2,

β3 : Z2 → Aut(Z4 × Z2),

β3(1) :

{(1, 0) 7→ (1, 0),(0, 1) 7→ (2, 1).

〈x, y, z | x4 = y2 = z2 = 1, xy = yx,

z−1xz = x, z−1yz = x2y〉

(Z4 × Z2) oβ4 Z2,

β4 : Z2 → Aut(Z4 × Z2),

β4(1) :

{(1, 0) 7→ (1, 1),(0, 1) 7→ (0, 1).

〈x, y, z | x4 = y2 = z2 = 1, xy = yx,

z−1xz = xy, z−1yz = y〉

Z4 oβ5 Z4

β5 : Z4 → Aut(Z4), [β1(1)](1) = −1.〈x, y | x4 = y4 = 1, y−1xy = x−1〉

Order 24. Table 1.3. (Cf. [1, §126].)

Order 27. Z9 oε1 Z3, ε1 : Z3 → Aut(Z9), [ε1(1)](1) = 4; (Z3 × Z3) oε2 Z3,

ε2 : Z3 → Aut(Z3 × Z3), ε2(1) :

{(1, 0) 7→ (1, 0),(0, 1) 7→ (1, 1).

Order 28. G ∼= D14 or Z7 o Z4.

Proof. Let P be the Sylow 7-subgroup of G and Q a Sylow 2-subgroup ofG. Then P ∼= Z7 and G ∼= P o Q. If Q ∼= Z4, G ∼= Z7 o Z4. If Q ∼= Z2 × Z2,G ∼= D14. �

Order 30.

Proof. Let P be the Sylow 5-subgroup of G and Q a Sylow 3-subgroup ofG. Since P C G, PQ < G and PQ ∼= Z5 × Z3. So, G ∼= (Z5 × Z3) oθ Z2, where

1.8. NONABELIAN GROUPS OF ORDER ≤ 30 23

Table 1.3. Nonabelian groups of order 24

group presentation

D4 × Z3

Q8 × Z3

Z3 oδ1 Z8,

δ1 : Z8 → Aut(Z3), [δ1(1)](1) = −1〈a, b | a8 = b3 = 1, a−1ba = b−1〉

Z3 oδ2 (Z4 × Z2),δ2 : Z4 × Z2 → Aut(Z3),

[δ2(1, 0)](1) = 1, [δ2(0, 1)](1) = −1

〈a, b, c | a3 = b4 = c2 = 1, bc = cb,

b−1ab = a, cac = a−1〉

Z3 oδ3 (Z4 × Z2),δ3 : Z4 × Z2 → Aut(Z3),

[δ3(1, 0)](1) = −1, [δ3(0, 1)](1) = 1

〈a, b, c | a3 = b4 = c2 = 1, bc = cb,

b−1ab = a−1, cac = a〉

Z2 ×D6

Z3 oδ4 (Z2 × Z2 × Z2)δ4 : Z2 × Z2 × Z2 → Aut(Z3),

[δ4(1, 0, 0)](1) = −1,[δ4(0, 1, 0)](1) = 1,[δ4(0, 0, 1)](1) = 1

〈a, b, c, d | a3 = b2 = c2 = d2 = 1,bc = cb, cd = dc, db = bd,

bab = a−1, cac = a, dad = a〉

Q8 oδ5 Z3,

δ5 : Z3 → Aut(Q8),

δ5(1) :

{i 7→ j

j 7→ k

〈a, b, c | a4 = c3 = 1, a2 = b2,

b−1ab = a−1, c−1ac = b, c−1bc = ab〉

Z3 oδ6 Q8,

δ6 : Q8 → Aut(Z3),[δ6(i)](1) = 1, [δ6(j)](1) = −1

〈a, b, c | a3 = b4 = 1, b2 = c2,

c−1bc = b−1, b−1ab = a, c−1ac = a−1〉

Z3 oδ7 D4,

δ7 : D4 → Aut(Z3),[δ7(α)](1) = 1, [δ7(β)](1) = −1

〈a, b, c | a3 = b4 = c2 = 1, cbc = b−1,

b−1ab = a, cac = a−1〉

Z3 oδ8 D4,

δ8 : D4 → Aut(Z3),[δ8(α)](1) = −1, [δ8(β)](1) = 1

〈a, b, c | a3 = b4 = c2 = 1, cbc = b−1,

b−1ab = a−1, cac = a〉

S4 〈a, b | a4 = b3 = (ab)2 = 1〉

θ : Z2 → Aut(Z5 × Z3),

θ(1) :

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0, 1),

or

{(1, 0) 7→ (1, 0),(0, 1) 7→ (0,−1),

or

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0, 1).

In the three cases, G ∼= D5 × Z3, S3 × Z5 and D15 respectively. �

24 1. GROUPS

Advanced Reading

1.9. Group Extensions

If 1 → Ki→ G

p→ Q → 1 is an exact sequence of groups, G is called anextension of K by G. If there is a homomorphism j : Q→ G such that p ◦ j = idQ,the exact sequence (or the extension) is called split. Given K,G,Q there exists

split sequence 1→ Ki→ G

p

�jQ→ 1 iff G ∼= K oG.

Q-modules induced by extensions. Let 0 → Ki→ G

p→ Q → 1 be anextension, where K is abelian and G is written additively. There exists a homo-morphism φ : Q→ Aut(K). Let j : Q→ G be any lifting. Then for x ∈ Q,

φ(x) : K −→ K

a 7−→ i−1(j(x) + i(a)− j(x)

).

φ is independent of j. φ defines an action of Q on K; hence K is a (left) Q-module.To recap, an extension 0→ K

i→ Gp→ Q→ 1 induces a Q-module K.

Given a Q-module K, an extension 0→ Ki→ G

p→ Q→ 1 is said to realize theQ-module K if the Q-module K coincides with the one induced by the extension.

Let K be a Q-module and f : Q×Q→ K a function. Define a binary operation+ on K ×Q:

(a, x) + (b, y) =(a+ xb+ f(x, y), xy

).

The (K ×Q,+) is a group ⇔

(1.5) xf(y, z)− f(xy, z) + f(x, yz)− f(x, y) = 0 ∀x, y, z ∈ Q.

f is called a 2-cocycle if (1.5) holds. (K ×Q,+) is denoted by G(K,Q, f) if f is a2-cocycle.

Note.(i) (1.5) ⇔ the associativity of +.(ii) (1.5) implies that f(1, y) = f(1, 1), f(x, 1) = xf(1, 1) ∀x, y ∈ Q. (1.5) also

implies the existence of identity and the inverse inG(K,Q, f). (0G(K,Q,f) =(−f(1, 1), 1), −(a, x) = (−x−1a− f(1, 1)− f(x−1, x), x−1).)

(iii) If f is a 2-cocycle, then there exists an exact sequence

(1.6) 0 −→ Ki−→ G(K,Q, f)

p−→ Q −→ 1

where

(1.7)i : K −→ G(K,Q, f)

a 7−→ (a− f(1, 1), 1),

(1.8)p : G(K,Q, f) −→ Q

(a, x) 7−→ x.

Moreover, extension (1.6) realizes the Q-module K.

Let 0 → Ki→ G

p→ Q → 1 be an extension making K a Q-module. Letj : Q→ G be a lifting. Then

(1.9) j∗(x, y) := i−1(j(x) + j(y)− j(xy)

)

1.9. GROUP EXTENSIONS 25

is a 2-cocycle. Moreover,

φ : G(K,Q, j∗) −→ G

(a, x) 7−→ i(a) + j(x)

is an isomorphism and the diagram

0 −→ Ki′−→ G(K,Q, j∗)

p′−→ Q −→ 1yid

yφ yid

0 −→ Ki−→ G

p−→ Q −→ 1

commutes, where i′ and p′ are defined by (1.7) and (1.8). If j′ : Q→ G is anotherlifting. Then j′(x)− j(x) = h(x) for some function h : Q→ K. Moreover,

j′∗(x, y)− j∗(x, y) = xh(y)− h(xy) + h(x).

The second cohomology group. Let K be a Q-module. Z2(Q,K) :=the abelian group of all 2-cocycles f : Q × Q → K. A function f : Q × Q → Kis called a 2-coboundary if f(x, y) = xh(y) − h(x, y) + h(x) for some h : Q →K. B2(Q,K) := the set of all 2-coboundaries. Then B2(Q,K) < Z2(Q,K).H2(Q,K) := Z2(Q,K)/B2(Q,K) is the 2nd cohomology group of Q with coeffi-cients in K.

Equivalence of extensions. Let K be a Q-module and let

0 −→ Ki−→ G

p−→ Q −→ 1(1.10)

0 −→ Ki′−→ G′

p′−→ Q −→ 1(1.11)

be two extensions realizing the Q-module K. The two extensions are called equiv-alent if there exists an isomorphism φ : G → G′ such that the following diagramcommutes.

0 −→ Ki−→ G

p−→ Q −→ 1yid

yφ yid

0 −→ Ki′−→ G′

p′−→ Q −→ 1

Theorem 1.28 (The meaning ofH2(Q,K)). Let K be a Q-module. Let E(Q,K)be the family of all equivalence classes of extensions of K by Q realizing the Q-module K. There is a bijection α : H2(Q,K)→ E(Q,K); α(0) = the class of splitextensions.

Proof. α : f + B2(Q,K) 7→ [G(K,Q, f)], f ∈ Z2(Q,K); α−1 : [G] 7→ j∗ +B2(Q,K), where j : Q→ G is a lifting and j∗ ∈ Z2(Q,K) is given by (1.9). �

Corollary 1.29. Let K be a Q-module. If H2(Q,K) = 0, then every exten-sion of K by Q is a semidirect product.

Proposition 1.30. Let K be a Q-module where |K| = m, |Q| = n and (m,n) =1. Then H2(Q,K) = 0.

Proof. Let f ∈ Z2(Q,K). Then

xf(y, z)− f(xy, z) + f(x, yz)− f(x, y) = 0 ∀x, y, z ∈ Q.Sum over z ∈ Q⇒

xh(y)− h(xy) + h(x) = nf(x, y),

26 1. GROUPS

where h(x) =∑z∈Q f(x, z). So, nf ∈ B2(Q,K). Since (m,n) = 1, f ∈ B2(Q,K).

�

Complement. Two subgroups H,K of G are called complements of each otherif G = HK and H ∩K = {0}.

Corollary 1.31. If |G| = mn, (m,n) = 1 and G has an abelian normalsubgroup K of order m. Then K has a complement.

Theorem 1.32 (The Schur-Zassenhaus theorem). If |G| = mn, (m,n) = 1 andG has a normal subgroup K of order m, then K has a complement.

Proof. Induction on m. If K has a nontrivial subgroup N such that N C G,K/N has a complement H/N in G/N . Since N C H, (|N |, |H|/|N |) = (|N |, n) = 1and |N | < m, H has a subgroup of order n.

So, we may assume that K is a minimal normal subgroup of G. Let P 6= 1be a Sylow subgroup of K. Since G = KNG(P ) (Frattini argument), G/K =KNG(P )/K ∼= NG(P )/K ∩ NG(P ) = NG(P )/NK(P ). If NG(P ) 6= G, then|NK(P )| < |K| = m. The induction hypothesis applied to NK(P ) C NG(P )implies that NG(P ) has a subgroup of order n.

So, assume NG(P ) = G, i.e. P C G. Then, 1 6= Z(P ) C G. Since Z(P ) ⊂ K,the minimality of K ⇒ K = Z(P ), which is abelian. Corollary 1.31 applies. �

1.10. Normal and Subnormal Series

Definition 1.33. Let G be a group.

(1.12) G = G0 B G1 B · · · B Gm = {e}

is called a subnormal series of G. Gi/Gi+1, 0 ≤ i ≤ m − 1, are the factor groupsof the series. The length of (1.12) is |{0 ≤ i ≤ m − 1 : Gi 6= Gi+1}|. If Gi C G,0 ≤ i ≤ m − 1, (1.12) is called a normal series. If Gi/Gi+1 is simple for all0 ≤ i ≤ m − 1, (1.12) is called a composition series. If Gi/Gi+1 is abelian for all0 ≤ i ≤ m− 1, (1.12) is called a solvable series.

A (sub)normal series

(1.13) G = H0 B H1 B · · · B Hn = {e}

is called a refinement of (1.12) if {G0, . . . , Gm} ⊂ {H0, . . . ,Hn}. Two subnormalseries S : G = G0 B · · · B Gm = {e} and T : G = H0 B · · · B Hn = {e} are calledequivalent if there is a bijection between the nontrivial factors of S and those of Tsuch that the corresponding factors are isomorphic.

Theorem 1.34 (The Schreier refinement theorem). Any two (sub)normal seriesof G have equivalent refinements.

Proof. Let S : G = G0 B · · · B Gm = {e} and T : G = H0 B · · · B Hn = {e}be two (sub)normal series. Let

Gij = Gi+1(Gi ∩Hj), Hij = Hj+1(Gi ∩Hj), 0 ≤ i ≤ m, 0 ≤ j ≤ n.

1.10. NORMAL AND SUBNORMAL SERIES 27

(Gm+1 := {e}, Hn+1 := {e}.) Then

Gi−1,n = Gi,0 = Gi, 1 ≤ i ≤ m− 1,Hm,j−1 = H0,j = Hj , 1 ≤ j ≤ n− 1,Gm−1,n−1 = Hm−1,n−1 = Gm−1 ∩Hn−1.

We have a refinement S ′ for S

S ′ :

G = G00 B G01 B · · · B G0,n−1 B

G10 B G11 B · · · B G1,n−1 B...

......

Gm−1,0 B Gm−1,1 B · · · B Gm−1,n−1 B {e}

and a refinement T ′ for T

T ′ :

G

qH00 H01 · · · H0,n−1

5 5 5H10 H11 · · · H1,n−1

5 5 5...

......

5 5 5Hm−1,0 Hm−1,1 · · · Hm−1,n−1

5 5 5{e}

It suffices to show Gij/Gi,j+1∼= Hij/Hi+1,j , 0 ≤ i ≤ m− 1, 0 ≤ j ≤ n− 1. By the

Zassenhaus lemma,

Gij/Gi,j+1 =Gi+1(Gi ∩Hj)Gi+1(Gi ∩Hj+1)

∼=Hj+1(Gi ∩Hj)Hj+1(Gi+1 ∩Hj)

= Hij/Hi+1,j .

�

Lemma 1.35 (Zassenhaus). Let A∗, A,B∗, B be subgroups of G such that A∗ CA and B∗ C B. Then A∗(A ∩B∗) C A∗(A ∩B), B∗(A∗ ∩B) C B∗(A ∩B) and

A∗(A ∩B)A∗(A ∩B∗)

∼=B∗(A ∩B)B∗(A∗ ∩B)

.

Proof. Let D = (A∗ ∩ B)(A ∩ B∗). We claim that D C A ∩ B. SinceA∗ C A, we have A∗ ∩ B C A ∩ B; since B∗ C B, we have A ∩ B∗ C A ∩ B. So,D = (A∗ ∩B)(A ∩B∗) C A ∩B.

It suffices to showA∗(A ∩B)A∗(A ∩B∗)

∼=A ∩BD

.

28 1. GROUPS

(By symmetry, we have B∗(A∩B)B∗(A∗∩B)

∼= A∩BD .) Define

φ : A∗(A ∩B) −→ A ∩BD

ax 7−→ Dx, a ∈ A∗, x ∈ A ∩B.

1◦ φ is well defined. If ax = a′x′, where a, a′ ∈ A∗ and x, x′ ∈ A ∩ B, thenx′x−1 = (a′)−1a ∈ A∗ ∩ (A ∩B) = A∗ ∩B ⊂ D. So, Dx = Dx′.

2◦ φ is onto. Obvious.3◦ φ is a homomorphism. ∀ ax, a′x′ ∈ A∗(A ∩ B), we have φ(axa′x′) =

φ(axa′x−1xx′) = Dxx′ = φ(ax)φ(a′x′) since xa′x−1 ∈ A∗.4◦ kerφ = A∗D = A∗(A∗ ∩B)(A ∩B∗) = A∗(A ∩B∗).By the 1st isomorphism theorem, A∗(A∩B)

A∗(A∩B∗)∼= A∩B

D . �

Fact. Every finite group has a composition series.

Theorem 1.36 (Jordan-Holder). Any two composition series of a group G areequivalent.

Proof. Let S and T be two composition series of G. By Theorem 1.34, S andT have refinements S ′ and T ′ such that S ′ and T ′ are equivalent. But S ′ = S andT ′ = T . �

Examples of composition series. Sn B An B {id}, where n ≥ 5. S4 BA4 B K B {id}.

Derived groups and solvable groups. Let G be a group. G(0) := G,G(i+1) := (G(i))′. G(i) is called the ith derived group of G. If G(n) = {e} for somen ∈ Z+, G is called solvable.

Proposition 1.37. G is solvable ⇔ G has a solvable series.

Proof. (⇒) G = G(0) B G(1) B · · · B G(n) = {e} is a solvable series.(⇐) Let G = G0 B G1 B · · · B Gn = {e} be a solvable series. Then Gi ⊃ G′i−1,

1 ≤ i ≤ n. So, {e} = Gn ⊃ G′n−1 ⊃ G′′n−2 ⊃ · · · ⊃ G(n)0 = G(n). �

Fact. Let N C G. Then G is solvable ⇔ N and G/N are both solvable.

Examples. Finite p-groups are solvable. Sn (n ≥ 5) is not solvable.

Theorem 1.38 (The Burnside p-q theorem). If |G| = paqb, where p, q areprimes, then G is solvable.

The proof needs representation theory.

Theorem 1.39 (The Feit-Thompson theorem). Every finite group of odd orderis solvable.

The proof is over 250 pages [6].

Central series and nilpotent groups.

{e} = G0 C G1 C · · · C Gn = G

such that Gi+1/Gi ⊂ Z(G/Gi) is called a central series of G. A central series is anascending series. If G has a central series, G is called nilpotent.

EXERCISES 29

Easy fact. Let G be a group. Z0(G) := {e}. Zi+1(G) is defined byZ(G/Zi(G)) = Zi+1(G)/Zi(G). Then G is nilpotent ⇔ Zn(G) = G for somen ∈ Z+.

Proof. (⇒) Use induction to show Gi ⊂ Zi(G). Assume Gi−1 ⊂ Zi−1(G).Since f : G/Gi−1 → G/Zi−1(G), gGi−1 7→ gZi−1(G), is an onto homomorphismand since Gi/Gi−1 ⊂ Z(G/Gi−1), we have GiZi−1(G)/Zi−1(G) = f(Gi/Gi−1) ⊂Z(G/Zi−1(G)) = Zi(G)/Zi−1(G). So, Gi ⊂ Zi(G). �

Easy fact. G nilpotent ⇒ G solvable.

Example. Let n = 2tm, 2 - m. Consider the dihedral group

Dn = 〈α, β | αn = β2 = 1, βαβ−1 = α−1〉.Then

(1.14) Zi(Dn) =

{〈αn/2i〉 if 0 ≤ i ≤ t,〈αm〉 if i > t.

(To see (1.14), use induction on i. Assume that i < t and Zi(Dn) = 〈αn/2i〉. ThenDn/Zi(Dn) = 〈α, β〉 ∼= Dn/2i , where α = αZi(Dn), β = βZi(Dn), o(α) = n/2i,o(β) = 2, βαβ−1 = α−1. Hence, Z(Dn/Zi(Dn)) = 〈αn/2i+1〉 = 〈αn/2i+1〉/Zi(Dn).So, Zi+1(Dn) = 〈αn/2i+1〉.) So, Dn is nilpotent iff n is a power of 2. (Note. Dn isalways solvable.)

Fact 1.40. If G is nilpotent and H � G, then H 6= NG(H).

Proof. Let {e} = G0 C G1 C · · · C Gn = G be a central series. Choosei such that Gi ⊂ H but Gi+1 6⊂ H. Since Gi+1/Gi ⊂ Z(G/Gi). Gi+1/Gi ⊂NG/Gi

(H/Gi)⇒ Gi+1 ⊂ NG(H). �

Fact. Finite p-groups are nilpotent.

Proposition 1.41. Let G be a finite group. Then G is nilpotent ⇔ G ∼= thedirect product of its Sylow subgroups.

Proof. (⇒) Only have to show that every Sylow subgroup P of G is nor-mal. By Proposition ??, NG(NG(P )) = NG(P ). If NG(P ) 6= G, by Fact 1.40,NG(NG(P )) 6= NG(P ), which is a contradiction. So, NG(P ) = G, i.e. P C G.

(⇐) H ×K is nilpotent ⇔ H and K are both nilpotent. �

Exercises

1.1. Let f : G → H be a homomorphism of groups and X ⊂ G. Prove thatf(〈X〉) = 〈f(X)〉.

1.2. (Double cosets) Let H and K be subgroups of G. For a ∈ G, HaK := {hak :h ∈ H, k ∈ K} is called an (H,K)-double coset in G.(i) Prove that the set of all (H,K)-cosets form a partition of G.(ii) Assume H and K are finite. Prove that |HaK| = |H||K|

|H∩aKa−1| .

1.3. (Cyclic groups)(i) Prove that every subgroup of a cyclic group is cyclic.(ii) Prove that every quotient group of a cyclic group is cyclic.

30 1. GROUPS

(iii) Let G be an infinite group. Prove that G ∼= Z ⇔ G is isomorphic toevery nontrivial subgroup of G.

(iv) If G is a cyclic group of order n. Then for every m | n, G has a uniquesubgroup of order m. (If G = 〈a〉, then 〈a n

m 〉 is the unique subgroup oforder m.)

1.4. (Diagram chasing) A sequence of groups and homomorphisms

· · · −→ Gi−1fi−1−→ Gi

fi−→ Gi+1 −→ · · ·is called exact at Gi if im fi−1 = ker fi; the sequence is called exact if it is

exact at Gi for all i. An exact sequence 1 → Kf→ G

h→ Q → 1 is called ashort exact sequence. Let

1 1 1y y y1 −→ A

f−→ Bg−→ C −→ 1

α

y yβ yγ1 −→ A′

f ′−→ B′ g′−→ C ′ −→ 1

α′y yβ′ yγ′

1 −→ A′′f ′′−→ B′′ g′′−→ C ′′ −→ 1y y y

1 1 1

be a commutative diagram of groups such that all columns are exact and the2nd and 3rd rows are exact. Prove that the 1st row is also exact.

1.5. Determine all subgroups of S4 and identify the ones that are normal.

1.6. (The converse of Lagrange’s theorem is false.) Prove that An does not havea subgroup of index 2.

1.7. Assume H < Sn such that [Sn : H] = 2. Prove that H = An.

1.8. (Application of Burnside’s lemma) Each vertex of a regular n-gon is to becolored with any of c colors. Two colored regular n-gons are considered thesame if one can be obtained from the other through a rotation or a reflection.Find the total number of different ways to color the regular n-gon. (Let Xbe the set of all colored regular n-gons. Then Dn acts on X and the numberto be found is the number of Dn-orbits in X .)

1.9. M1 Prove that there are no simple groups of order 120 and 300.

1.10. (Groups of order 2pq) Let G be a nonabelian group of order 2pq, where2 < p < q are primes.(i) Assume q 6≡ 1 (mod p). Then

G ∼= (Zq × Zp) oαiZ2

for some i = 1, 2, 3, where αi : Z2 → Aut(Zq × Zp),

α1(1) :

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0, 1);

α2(1) :

{(1, 0) 7→ (1, 0),(0, 1) 7→ (0,−1);

α3(1) :

{(1, 0) 7→ (−1, 0),(0, 1) 7→ (0,−1).

EXERCISES 31

(ii) Assume q ≡ 1 (mod p). In addition to the three groups in (i), there isa fourth group

(Zq o Zp) oβ Z2,

where Zq o Zp = 〈a, b | aq = bp = 1, bab−1 = ak〉, o(k) = p in Z×p , andβ : Z2 → Aut(Zq o Zp),

β(1) :

{a 7→ a−1,

b 7→ b.

Note that

(Zq o Zp) oβ Z2 = 〈a, b, c | aq = bp = c2 = 1, bab−1 = ak, cac−1 = a−1, cbc−1 = b〉.

CHAPTER 2

Rings and Modules

2.1. Rings, Basic Definitions

Definition 2.1. A ring is a nonempty set R equipped with two operations +and · such that

(i) (R,+) is an abelian group;(ii) (ab)c = a(bc) ∀a, b, c ∈ R;(iii) a(b+ c) = ab+ ac, (a+ b)c = ac+ bc ∀a, b, c ∈ R.

If ab = ba for all a, b ∈ R, R is called commutative. If ∃1R ∈ R such that 1Ra =a1R = a ∀a ∈ R, 1R is called the identity of R.

Subring. Let (R,+, ·) be a ring. S ⊂ R is called a subring of R if (S,+, ·) isa ring.

Homomorphism. Let R and S be rings. A map f : R → S is called ahomomorphism if f(a + b) = f(a) + f(b), f(ab) = f(a)f(b) for all a, b ∈ R. Anisomorphism is a bijective homomorphism.

Note. In general, a ring may not have an identity, e.g. 2Z. If S is a subringof R, any of the following could happen: (i) R has identity, S does not (R = Z,S = 2Z); (ii) S has identity, R does not (R = Z × 2Z, S = Z × {0}); (iii) R andS both have identity but 1R 6= 1S (R = Z× Z, S = Z× {0}). If R and S are tworings with identity, a homomorphism f : R → S does not necessarily map 1R to1S . However, we make the following declaration.

Declaration. In these notes, unless specified otherwise, it is assumed that aring has identity; if S is a subring of R, 1S = 1R; a homomorphism maps identityto identity.

Basic properties of rings.(i) 0R · a = a · 0R = 0R, a ∈ R.(ii) (na)b = a(nb) = n(ab), m(na) = (mn)a, a, b ∈ R, m,n ∈ Z.(iii) ( n∑

i=1

ai

)( m∑j=1

bj

)=

n∑i=1

m∑j=1

aibj .

(iv) Assume a1, . . . , as ∈ R are pairwise commutative. Then

(a1 + · · ·+ as)n =∑

i1+···+is=n

n!i1! · · · is!

ai11 · · · aiss .

The multiplicative group. a ∈ R is call a unit (or invertible) if ∃b ∈ R suchthat ab = ba = 1R. R× := the set of all units of R. (R×, ·) is the multiplicativegroup of R.

33

34 2. RINGS AND MODULES

Types of rings.Integral domain. R: commutative, 1R 6= 0, no zero divisors (i.e., ab = 0⇒ a =

0 or b = 0).Division ring (skew field). R: 1R 6= 0, R× = R \ {0}.Field. Commutative division ring.

Examples.Fields: Q, R, C, Zp (p prime).Integral domains (not fields): Z, D[x] (the polynomial ring over an integral

domain D).Noncommutative rings: Mn×n(R) = the ring of n× n matrices over a ring R.

Endomorphism ring. Let A be an abelian group, End(A) = Hom(A,A).(End(A),+, ◦) is the endomorphism ring of A.

Fact. Every ring R is a subring of End((R,+)

).

Proof. We havef : R ↪→ End

((R,+)

)r 7−→ f(r)

wheref(r) : (R,+) −→ (R,+)

x 7−→ rx.

�

Example (Real quaternions, a division ring which is not a field).

H = {a1 + a2i+ a3j + a4k : a1, . . . , a4 ∈ R}.Addition: coordinate wise; multiplication: defined by the distributive laws and therules i2 = j2 = k2 = −1, ij = k, jk = i, ki = j, ik = −j, kj = −i, ji = −k. Ifz = a1 + a2i+ a3j + a4k, define z = a1 − a2i− a3j − a4k. zz = a2

1 + a22 + a2

3 + a24.

If z 6= 0, z−1 = 1zz z.

Group rings. Let G be a group (written multiplicatively) and R a ring. Thegroup ring R[G] := the set of all formal sums

∑g∈G rgg, where rg ∈ R and rg = 0

except for finitely many g ∈ G.∑g∈G

rgg +∑g∈G

sgg :=∑g∈G

(rg + sg)g,(∑h∈G

rhh)(∑

k∈G

skk)

=∑g∈G

( ∑h,k∈Ghk=g

rhsk

)g.

If X ⊂ G is closed under multiplication and e ∈ X, then R[X] = {∑g∈X rgg ∈

R[G]} is a subring of R[G].

Characteristic. The characteristic of a ring R (charR) is the smallest n ∈Z+ such that na = 0 for all a ∈ R. If no such n exists, charR = 0. ( char Zn = n,char Q = 0.)

Fact. If D is an integral domain, charD = 0 or a prime.

Ideals. Let R be a ring. I ⊂ R is called a left (right) ideal of R if I is asubgroup of (R,+) and ax ∈ R for all a ∈ R, x ∈ I. An ideal is a two-sided ideal.

2.2. PRIME IDEALS AND MAXIMAL IDEALS 35

If X ⊂ R, the ideal of R generated by X (the smallest ideal containing X) is

〈X〉 (or (X)) ={ n∑i=1

aixibi : n ≥ 0, ai, bi ∈ R, xi ∈ X}.

Sum and product of ideals. Let I, J be left (right) ideals of R. Define

I + J = {a+ b : a ∈ I, b ∈ J}.I + J is the smallest left (right) ideal of R containing I ∪ J .

If I and J are ideals of R, define

IJ ={ n∑i=1

aibi : n ≥ 0, ai ∈ I, bi ∈ J}.

IJ is an ideal of R and IJ ⊂ I ∩ J .

The quotient ring. Let I be an ideal of R. Then R/I is an abelian group.For a + I, b + I ∈ R/I, define (a + I)(b + I) = ab + I. The multiplication is welldefined and (R/I,+, ·) is a ring, called the quotient ring of R by I.

π : R −→ R/I

r 7−→ r + I

is an onto homomorphism (canonical homomorphism).

Fact. I is an ideal of R ⇔ I = ker f for some homomorphism f : R→ S.

Proposition 2.2 (Universal mapping property). Let f : R → S be a homo-morphism of rings and let I be an ideal of R such that I ⊂ ker f . Then there existsa unique homomorphism f : R/I → S such that the following diagram commutes.

................................................................................................. .......................................................................................... ............................

........................................................

.........................................

f

πf

R S

R/I

Isomorphism theorems.(i) Let f : R→ S be a homomorphism of rings. Then R/ ker f ∼= f(R).(ii) Let I ⊂ J be ideals of R. Then (R/I)/(J/I) ∼= R/J .

The correspondence theorem. Let I be an ideal of R. Let A = the setof all ideals of R containing I, B = the set of all ideals of R/I. Then A → B:J 7→ J/I, is a bijection.

2.2. Prime Ideals and Maximal Ideals

Definition 2.3. An ideal P of R is called a prime ideal if (i) P 6= R and (ii)if A,B are ideals of R such that AB ⊂ P , then A ⊂ P or B ⊂ P .

An ideal M of R is called maximal if M 6= R and there is no ideal strictlybetween M and R. Maximal left (right) ideals are defined in the same way.

Proposition 2.4. Let P be an ideal of R such that P 6= R.(i) If for all a, b ∈ P , ab ∈ P implies a ∈ P or b ∈ P , then P is prime.(ii) If R is commutative, the converse of (i) is true.

36 2. RINGS AND MODULES

Proof. (i) Suppose AB ⊂ P and A 6⊂ P . Choose a ∈ A \ P . For all b ∈ B,ab ∈ AB ⊂ P . So, b ∈ P ; hence B ⊂ P .

(ii) Assume ab ∈ P . Then (a)(b) = (ab) ⊂ P ⇒ (a) ⊂ P or (b) ⊂ P . �

Note. If R is not commutative, the converse of (i) is false. Example: R =M2×2(F ) where F is any field. The only ideals of R are 0 and R. So, 0 is a primesideal of R. But [ 1

0 ][ 01 ] = 0.

Proposition 2.5. Let R be a ring and I 6= R a (left) ideal of R. Then I iscontained in a maximal (left) ideal of R.

Proof. Look at all (left) ideals J such that I ⊂ J 63 1. Use Zorn’s lemma. �

Theorem 2.6. Let R be a commutative ring and I an ideal of R.

(i) I is prime ⇔ R/I is an integral domain.(ii) I is maximal ⇔ R/I is a field.(iii) I is a maximal ⇒ I is prime.

Fact. If I is an ideal of a ring R such that R/I is a division ring, then I is amaximal ideal. The converse is false: 0 is a maximal ideal of M2×2(F ).

Proposition 2.7. Let I1, . . . , In be ideals of R such that I1 + · · ·+ In = R andIiIj = {0} for all i 6= j. Write 1 = e1 + · · ·+ en, where ei ∈ Ii. Then we have thefollowing conclusions.

(i)

eiej =

{ei if i = j,

0 if i 6= j.

(e1, . . . , en are called orthogonal idempotents.)(ii) Ii is a ring with identity ei. (It follows that e1, . . . , en are unique.) More-

over, e1, . . . , en are in the center of R and Ii = eiRei.(iii) R ∼= I1 × · · · × In.

Proof. (i) If i 6= j, then eiej ∈ IiIj = {0}; hence eiej = 0. Thus ei =ei(e1 + · · ·+ en) = e2i .

(ii) Let x ∈ Ii. Then for each j 6= i, xej ∈ IiIj = {0}; hence xej = 0. So,x = x(e1 + · · ·+ en) = xei. In the same way, eix = x.

Since ei is the identity of Ii and eix = 0 = xei for all x ∈ Ij , j 6= i, we see thatei is in the center if R. Since eiRei ⊂ Ii ⊂ eiIiei ⊂ eiRei, we have Ii = eiRei.

(iii) f : R → I1 × · · · × In, a 7→ (ae1, . . . , aen) is an isomorphism. (In fact,g : I1 × · · · × In → R, (x1, . . . , xn) 7→ x1 + · · ·+ xn, is the inverse of f .) �

Theorem 2.8 (The Chinese remainder theorem). Let I1, . . . , In be ideals of aring R such that Ii + Ij = R (i 6= j). Then

f : R −→ (R/I1)× · · · × (R/In)a 7−→ (a+ I1, . . . , a+ In)

is an onto homomorphism with ker f = I1 ∩ · · · ∩ In. (I.e., ∀ai ∈ Ii, 1 ≤ i ≤ n,∃a ∈ R (unique mod I1 ∩ · · · ∩ In) such that a ≡ ai (mod Ii) for all 1 ≤ i ≤ n.)

2.3. FACTORIZATION IN COMMUTATIVE RINGS; UFD, PID AND ED 37

Proof. Only have to show that f is onto. It suffices to show that ∃a ∈ R suchthat

a ≡

{1 (mod I1),0 (mod Ii), 2 ≤ i ≤ n.

Since I1 + Ii = R (i ≥ 2), there exists ai ∈ I1 such that ai ≡ 1 (mod Ii). Thena = (1− a2) · · · (1− an) works. �

Corollary 2.9. Let m1, . . . ,mn ∈ Z+ such that (mi,mj) = 1, i 6= j. Letai, . . . , an ∈ Z be arbitrary. Then there exists x ∈ Z (unique mod lcm(m1, . . . ,mn))such that x ≡ ai (mod mi) for all 1 ≤ i ≤ n.

Example. Let X be a compact topological space and C(X,R) the ring of allcontinuous functions from X to R. For each a ∈ X, let Ma = {f ∈ C(X,R) :f(a) = 0}. Then Ma, a ∈ X, are all the maximal ideals of C(X,R).

Proof. C(X,R)/Ma∼= R is a field. So, Ma is maximal.

Let M be a maximal ideal of C(X,R). Assume to the contrary that M 6= Ma

for all a ∈ X. Then ∀a ∈ X, ∃fa ∈ C(X,R) such that fa(a) 6= 0. So, fa(x)2 > 0 forall x in an open neighborhood Ua of a. Let Ua1 , . . . , Uan

be a finite cover ofX. Thenf2a1

+ · · ·+ f2an∈M is invertible. So, M = C(X,R), which is a contradiction. �

2.3. Factorization in Commutative Rings; UFD, PID and ED

Let R be a commutative ring and a, b ∈ R. a | b (a divides b) means thatb = ax for some x ∈ R. If a | b and b | a, then a, b are called associates, denotedas a ∼ b. (If R is an integral domain, a ∼ b ⇔ a = bu for some u ∈ R×.) Anelement a ∈ R \ (R× ∪ {0}) is called irreducible if a = bc (b, c ∈ R) ⇒ b or c is aunit. a ∈ R \ (R× ∪ {0}) is called prime if a | bc (b, c ∈ R) ⇒ a | b or a | c.

Definition 2.10 (PID). An integral domain P is called a principal ideal do-main (PID) if every ideal of P is principal.

Definition 2.11 (UFD). An integral domain R is called a unique factorizationdomain (UFD) if

(i) ∀a ∈ R \ (R× ∪ {0}), a = c1 · · · cn for some irreducible c1, . . . , cn ∈ R;(ii) if c1 · · · cn = d1 · · · dm, where ci, dj ∈ R are irreducible, then n = m and

after a suitable reordering, ci ∼ di, 1 ≤ i ≤ n.

Definition 2.12 (ED). An integral domain R is called a Euclidean domain(ED) if ∃∂ : R \ {0} → N such that

(i) ∀a, b ∈ R \ {0}, ∂(a) ≤ ∂(ab);(ii) ∀a ∈ R, 0 6= b ∈ R, ∃ q, r ∈ R such that a = qb + r, where r = 0 or

∂(r) < ∂(b).

Note.

(i) If ∂ satisfies (i) and (ii) of Definition 2.12, so does ∂ − min{∂(x) : x ∈R \ {0}}. Thus, we may assume 0 is in the range of ∂.

(ii) Let R be an ED and 0 6= x ∈ R. Then x ∈ R× ⇔ ∂(x) = min{∂(y) : y ∈

R \ {0}}.

Proposition 2.13. Let R be an integral domain.(i) p ∈ R is prime ⇔ (p) is a nonzero prime ideal.

38 2. RINGS AND MODULES

(ii) a ∈ R is irreducible ⇔ (a) is maximal in {(b) : 0 6= b ∈ R, (b) 6= R}.(iii) p is prime ⇒ p is irreducible.(iv) If R is a UFD, p is a prime ⇔ p is irreducible.

Proof. (iii) Suppose p = ab. Then p | ab ⇒ p | a (say). So, a = pu (u ∈ R),p = pub ⇒ ub = 1⇒ b is a unit.

(iv) (⇐) Assume p | ab (a, b ∈ R). Then pq = ab for some q ∈ R. By theuniqueness of factorization, p appears in the factorization of a or b, i.e., p | a orp | b. �

Note. If R is not a UFD, p irreducible 6⇒ p prime. Example: R = Z[√−5] :=

{a + b√−5 : a, b ∈ Z}. 2 ∈ R is irreducible. (If 2 = xy for some x, y ∈ Z[

√−5].

Then 4 = |2|2 = |x|2|y|2. It follows that of |x|2 and |y|2, say |x|2, is 1; hence x isinvertible.) 2 | 6 = (1 +

√−5)(1−

√−5). But 2 - (1 +

√−5), 2 - (1−

√−5).

Fact. ED ⇒ PID ⇒ UFD.

Proof. ED ⇒ PID. Let R be an ED and I 6= {0} an ideal of R. Let a ∈ Isuch that ∂(a) is the smallest. Then I = (a).

PID ⇒ UFD.Existence of factorization. Let a ∈ R\(R×∪{0}). Assume to the contrary that

a is not a product of finitely many irreducibles. Since a is not irreducible, a = a1a′1,

where a1, a′1 ∈ R \ (R× ∪ {0}) and w.l.o.g., a1 is not a product of finitely many

irreducibles. Write a1 = a2a′2, ... ⇒ (a) ( (a1) ( (a2) ( · · · .

⋃∞i=1(ai) is an ideal of

R. So,⋃∞i=1(ai) = (b) for some b ∈ R⇒ b ∈ (ai) for some i ⇒ (ai+1) ⊂ (b) ⊂ (ai),

which is a contradiction.Uniqueness of factorization. First show that every irreducible element a of R

is a prime. (By Proposition 2.13 (ii), (a) is a maximal ideal; hence (a) is a primeideal and a is a prime.) Then use induction on the number of irreducible factors inthe factorization. �

Examples of ED. Z, F [x] (F a field), and (cf. [8, §5.4])

Z[√d], d = −2, −1, 2, 3, 6, 7, 11, 19,

Z[ 1+√d

2 ], d = −11, −7, −3, 5, 13, 17, 21, 29, 33, 37, 41, 57, 73.

Example (UFD 6⇒ PID). Z[x]. (2, x) is not a principal ideal.

Example 2.14 (PID 6⇒ ED). Z[α], α = 12 (1 +

√−19).

Proof. 1◦ Z[α] is not a ED.The units of Z[α] are ±1. (u ∈ Z[α] is a unit ⇔ |u|2 = 1.) Assume to the

contrary that Z[α] is an ED with degree function ∂. We may assume that 0 ∈ im ∂.Let ε ∈ Z[α] such that ∂(ε) is the smallest in Z+. We have

2 = qε+ r, r = 0,±1.

So, qε = 1, 2, 3. Thus |ε|2∣∣ 12, 22, 32 ⇒ |ε|2 = 1, 2, 4, 3, 9. Also,

α = q1ε+ r1, r1 = 0,±1.

So, q1ε ∈ 12

√−19 + 1

2{±1, 3} ⇒ |ε|2∣∣ 1

4 (19 + 12) or 14 (19 + 32), i.e. |ε|2

∣∣ 5 or 7. So,|ε|2 = 1, which is a contradiction.

2◦ ∀z ∈ C, ∃q ∈ Z[α] such that either |z − q| < 1 or |z − q2 | <

12 .

2.3. FACTORIZATION IN COMMUTATIVE RINGS; UFD, PID AND ED 39

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........α α+1

12

1

α2

0• •

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◦ ◦ ◦

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•

◦

α

α2

0

Figure 2.1. Example 2.14

Let z = x+ yi. ∃p ∈ Z[α] such that z+ p belongs to the (closed) parallelogram0, 1, α + 1, α, see Figure 2.1. We want to show that z has distance < 1 from oneof the dots or has distance < 1

2 from one of the circles. For this purpose, we mayassume z ∈ 4(0, 1

2 , α). Assume |z − α2 | ≥

12 ⇒ (x − 1

4 )2 + (y −√

194 )2 ≥ 1

4 . Since|x − 1

4 | ≤14 , we have |y −

√194 | ≥

√3

4 ⇒ y ≤√

19−√

34 or y ≥

√19+

√3

4 . In the firstcase, |z − 0| < 1; in the second case, |z − α| < 1.

3◦ Z[α] is a PID.Let I 6= {0} be an ideal of Z[α]. Let 0 6= β ∈ I such that |β|2 is the smallest.

We claim that I = (β).∀σ ∈ I, by 2◦, ∃q ∈ Z[α] such that |σβ − q| < 1 or |σβ −

q2 | <

12 . If |σβ − q| < 1,

then |σ − qβ| < |β| ⇒ σ − qβ = 0 ⇒ σ ∈ (β). So, assume |σβ −q2 | <

12 . Then

|2σ − qβ| < |β| ⇒ σ = q2β. It suffices to show that q

2 ∈ Z[α]. Assume the contrary.Then q = a+ bα, where at least one of a, b is odd.

(i) a is odd, b is even. Then q+12 ∈ Z[α]⇒ 1

2β = σ− q+12 β ∈ I with 0 < | 12β| <

|β|, contradiction.(ii) a is even, b is odd. We have

qα = aα+ 5b = (a+ 5b)− aα = a′ + b′α =: q′,

where q′

2 β ∈ I, a′ odd, b′ even. This is (i).

(iii) a, b both odd. We have

qα = (a+ 5b)− aα = a′ + b′α =: q′,

where q′

2 β ∈ I, a′ even, b′ odd. This is (ii). �

Gauss integers. Z[i] := {a+ bi : a, b ∈ Z} is an ED with ∂(α) = |α|2.

Proof. Let α, β ∈ Z[i], β 6= 0. ∃ q ∈ Z[i] such that |αβ − q| < 1. So, |α−βq| <|β|. �

40 2. RINGS AND MODULES

Primes in Z[i]. Let α ∈ Z[i] be neither 0 nor a unit. Then α is a prime (i.e.irreducible) ⇔

(i) α ∼ p for some prime p ∈ Z with p ≡ −1 (mod 4) or(ii) |α|2 is prime in Z.

Proof. (⇐) Assume (i). Assume to the contrary that p is not a prime. ⇒p = βγ, where β, γ ∈ Z[i], |β|2 > 1, |γ|2 > 1. Since p2 = |β|2|γ|2 (in Z) ⇒ p =|β|2 ⇒ p 6≡ −1 (mod 4), →←.

Assume (ii). If α = βγ, where β, γ ∈ Z[i], ⇒ |α|2 = |β|2|γ|2 (in Z) ⇒ |β|2 = 1or |γ|2 = 1.

(⇒) We have |α|2 = p1 · · · pn, where p1, . . . , pm are primes in Z. Since α | αα =p1 · · · pn and α is prime, α | pi =: p for some i. So, |α|2

∣∣ p2 in Z, ⇒ |α|2 = p

or p2. If |α|2 = p, we have (ii). So, assume |α|2 = p2. Since α | p, p = uα forsome u ∈ Z[i]. So, |u|2 = 1, i.e., u is a unit. It remains to show that p ≡ −1(mod 4). If p = 2 or p ≡ 1 (mod 4), by Lemma 2.15, p = a2 + b2 for some a, b ∈ Z,⇒ α = u−1p = u−1(a+ bi)(a− bi) is not irreducible, which is a contradiction. �

Lemma 2.15. Let p be an odd prime integer. Then the following are equivalent.(i) p ≡ 1 (mod 4).(ii) −1 is a square in Zp.(iii) p = a2 + b2 for some a, b ∈ Z.

Proof. (i) ⇒ (ii). 4 | p− 1 = |Z×p | ⇒ ∃x ∈ Z×p with o(x) = 4 ⇒ −1 = x2.(ii) ⇒ (iii). We claim that p is not irreducible in Z[i]. (Otherwise, by (ii),

∃x ∈ Z such that p | x2 + 1 = (x + i)(x − i) ⇒ p | x + i or p | x − i ⇒ x ± i =p(a + bi) ⇒ ±1 = pb, contradiction.) So, p = αβ, where α, β ∈ Z[i] are nonunits,⇒ p2 = |α|2|β|2 (in Z) ⇒ p = |α|2 or |β|2. �

Theorem 2.16 (Sum of two squares). Let n ∈ Z+ have factorization n =pe11 · · · pem

m qf11 · · · qfnn , where p1, . . . , pm, q1, . . . , qn are distinct primes with pi ≡ −1

(mod 4) and qj = 2 or qj ≡ 1 (mod 4). Then n = a2 + b2 for some a, b ∈ Z ⇔e1, . . . , em are all even.

Proof. (⇐) qj = |αj |2 for some αj ∈ Z[i],⇒ n = |pe1/21 · · · pem/2m αf11 · · ·αfn

n |2.(⇐) We have n = αα for some α ∈ Z[i]. Assume to the contrary that ei is odd

for some i. Write ei = 2k + 1. Since pi is a prime of Z[i] and p2k+1i | αα, we have

pk+1i | α or α. Then pk+1

i | α and α. So, p−ei−1i n =

∣∣∣ α

pk+1i

∣∣∣2 ∈ Z, →←. �

gcd and lcm. Let R be a commutative ring and X ⊂ R. An element d ∈ Ris called a greatest common divisor of X, denoted by gcd(X), if

(i) d | x ∀x ∈ X and(ii) if c | x ∀x ∈ X, then c | d.

An element m ∈ R is called the least common multiple of X, denoted by lcm(X),if

(i’) x | d ∀x ∈ X and(ii’) if x | c ∀x ∈ X, then m | c.

gcd’s (lcm’s) of X may not exist. If they do, all gcd’s (lcm’s) of X are associates.If R is a PID, then 〈gcd(X)〉 = 〈X〉 and 〈lcm(X)〉 =

⋂x∈X〈x〉.

2.4. FRACTIONS AND LOCALIZATION 41

Assume R is a UFD. Two primes in R which are associates will be treatedas being the same. Let P be the set of all distinct primes in R. Then for eachx ∈ R \ {0},

x ∼∏p∈P

pνp(x),

where νp(x) ∈ N and νp(x) = 0 for almost all p ∈ P. Also define νp(0) = ∞ forall p ∈ P. Moreover, define

∏p∈P p

ep = 0 if ep = ∞ for some p ∈ P or ep > 0 forinfinitely many p ∈ P. Then

gcd(X) ∼∏p∈P

pinf{νp(x):x∈X},

lcm(X) ∼∏p∈P

psup{νp(x):x∈X}.

2.4. Fractions and Localization

The ring of fractions. Let R be a commutative ring and let ∅ 6= S ⊂ R\{0}be a multiplicative set (i.e., S is closed under multiplication). For (r, s), (r′, s′) ∈R × S, define (r, s) ∼ (r′, s′) if ∃s1 ∈ S such that s1(rs′ − r′s) = 0. “∼” is anequivalence relation on R × S. The equivalence class of (r, s) in R × S is denotedby r

s . Let S−1R = R× S/ ∼= { rs : r ∈ R, s ∈ S}. For rs ,

r′

s′ ∈ R, define

r

s· r

′

s′=rr′

ss′,

r

s+r′

s′=rs′ + sr′

ss′.

Then (S−1R,+, ·) is a commutative ring, called the ring of fractions of R by S. IfR is an integral domain, so is S−1R. If R is a integral domain and S = R \ {0},S−1R is a field, called the fractional field of R.

Examples. Q = the fractional field of Z. The fractional field of F [x] (F afield) is F (x), the field of rational functions over F .

Proposition 2.17. Let R be a commutative ring and S (6= ∅, 63 0) a multipli-cation set of R.

(i) The map

φS : R −→ S−1R

r 7−→ rss (s ∈ S arbitary)

is a homomorphism. For every s ∈ S, φS(s) is a unit of S−1R.(ii) φS is 1-1 ⇔ S contains no zero divisors.

Proposition 2.18 (Universal mapping property). Let R be a commutativering and S (6= ∅, 63 0) a multiplication set of R. Let T be another commutative ringand f : R → T a homomorphism such that f(S) ⊂ T×. Then there is a uniquehomomorphism f : S−1R→ T such that the following diagram commutes.

................................................................................................. .......................................................................................... ............................

........................................................

.........................................

f

φSf

R T

S−1R

42 2. RINGS AND MODULES

Proof. Existence. Define f : S−1R→ T , rs 7→ f(r)f(s)−1.Uniqueness. Assume g : S−1R → T is another homomorphism such that

g ◦ φS = f . Then for each r ∈ R and s ∈ S, g( rs )f(s) = g( rs )g(s2

s ) = g( rs2

s2 ) = f(r);hence g( rs ) = f(r)f(s)−1. �

Local rings. A local ring is a commutative ring R with a unique maximalideal M . R/M is called the residue field of R. Example: Let p be a prime andn > 0. Zpn is a local ring with maximal ideal pZpn and residue field Zpn/pZpn ∼= Zp.

Proposition 2.19. Let R be a commutative ring.(i) If R is local, the unique maximal ideal of R is R \R×.(ii) R is local ⇔ R \R× is closed under +.

Proof. (i) Let M be the unique ideal of R. ∀x ∈ R \ R×, by Zorn’s lemma,x ∈M . So, R \R× ⊂M . Clearly, M ⊂ R \R×. So, M = R \R×.

(ii) (⇐) R \ R× is an ideal of R. Let M be any maximal ideal of R. ThenM ⊂ R \R×. Hence M = R \R× is unique. So, R is local. �

Localization. Let R be a commutative ring and P a prime ideal of R. ThenS = R \ P is multiplicative subset of R and 0 /∈ S 6= ∅. S−1R is a local ring withmaximal ideal S−1P . ( If rs ∈ (S−1R) \ (S−1P ), then r ∈ R \P . So, rs is invertiblein S−1R.) S−1R is called the localization of R at P and denoted by RP . Example:Let p ∈ Z be a prime. Then Z(p) = {ab : a, b ∈ Z, p - b}.

2.5. Polynomial Rings

Polynomial ring in one indeterminate. Let R be a ring. A polynomialin x (the indeterminate) with coefficients in R is a formal sum

f = a0 + a1x+ · · ·+ anxn, n ∈ N, ai ∈ R.

deg f := max{i : ai 6= 0}. (deg 0 = −∞.) R[x] := the set of all polynomials in xwith coefficients in R. + and · in R[x] are defined as follows:

n∑i=0

aixi +

n∑i=0

bixi =

n∑i=0

(ai + bi)xi;

( n∑i=0

aixi)( m∑

j=0

bjxj)

=n+m∑k=0

( ∑i+j=k

aibj

)xk.

(R[x],+, ·) is a ring, called the polynomial ring over R in x.

Polynomial ring in a set of indeterminates. Let R be a ring. Let Xbe a set of symbols (indeterminates). Let A be the set of all functions α : X → Nsuch that α(x) = 0 for almost all (all but finitely many) x ∈ X. A polynomial inX with coefficients in R is a formal sum

f =∑α∈A

aαXα,

where aα = 0 for almost all α ∈ A. If x1, . . . , xn ∈ X (distinct) and α ∈ A suchthat

α(x) =

{di if x = xi, 1 ≤ i ≤ n,0 if x ∈ X \ {x1, . . . , xn},

2.5. POLYNOMIAL RINGS 43

we write Xα = xd11 · · ·xdnn . R[X] := the set of all polynomials in X with coefficients

in R. + and · in R[X] are defined as follows:∑α∈A

aαXα +

∑α∈A

bαXα =

∑α∈A

(aα + bα)Xα;(∑α∈A

aαXα)(∑

β∈A

bβXβ)

=∑γ∈A

( ∑α+β=γ

aαbβ

)Xγ .

(R[X],+, ·) is the polynomial ring over R in X.

Note. Let F be the free abelian group on X (written multiplicatively) and

X = {xd11 · · ·xdnn : n ≥ 0, xi ∈ X, di ∈ Z+}.

Then X is a multiplicative set of F containing 1. The subring R[X ] of the groupring R[F ] is precisely the polynomial ring R[X].

Note. ∀f ∈ R[X], ∃x1, . . . , xn ∈ X such that f ∈ R[x1, . . . , xn].

Proposition 2.20 (Universal mapping property). Let R[X] be the polynomialring over R in X. Let S be another ring and f : R → S a homomorphism. Letφ : X → S be a function such that every element in φ(X) commutes with everyelement in φ(X) ∪ f(R). Then there exists a unique homomorphism f : R[X]→ Ssuch that the following diagram commutes.

.........................................................................................................................................................................................................

...............................................................................................................................................

................................................................................................. ............................................................................................................

R

R[x]

S

X

f

f φ

.................

.................

Proof. Define f : R[X]→ S by∑d1,...,dn

ad1,...,dnxd11 · · ·xdn

n 7→∑

d1,...,dn

f(ad1,...,dn)φ(x1)d1 · · ·φ(xn)dn .

�

Fact. If X and Y are disjoint, then (R[X])[Y ] ∼= R[X ∪ Y ].

Proof. By Proposition 2.20, ∃ homomorphism g : (R[X])[Y ]→ R[X∪Y ] suchthat g|R[X] = id and g|Y = id. Also, ∃ homomorphism h : R[X ∪ Y ] → (R[X])[Y ]such that h|R[X] = id and h|Y = id. By the uniqueness of Proposition 2.20, g◦h = idand h ◦ g = id. �

Proposition 2.21 (The division algorithm). Let R be a ring and f, g ∈ R[x]such that the leading coefficient of g is a unit. Then ∃!q, r, q′r′ ∈ R[x] such that

f = qg + r and f = gq′ + r′,

where deg r < deg g, deg r′ < deg g.

Fact. If F is a field, F [x] is a ED with ∂(f) = deg f .

Let R be a commutative ring, f =∑d1,...,dn

ad1,...,dnxd11 · · ·xdn

n ∈ R[x1, . . . , xn]and (c1, . . . , cn) ∈ Rn. We write f(c1, . . . , cn) =

∑d1,...,dn

ad1,...,dncd11 · · · cdn

n . Iff(c1, . . . , cn) = 0, (c1, . . . , cn) is called a root of f .

44 2. RINGS AND MODULES

Facts.

(i) Let R be a commutative ring, f ∈ R[x] and c ∈ R. Then f(c) = 0 ⇔x− c | f .

(ii) If D is an integral domain and 0 6= f ∈ D[x] with deg f = n, then f hasat most n distinct roots in D.

Derivative. Let R be a commutative ring and f = a0 + · · · + anxn ∈ R[x].

f ′ := a1 + 2a2x+ · · ·+ nanxn−1. The differentiation rules hold.

The Multiplicity of a root. Let R be a commutative ring, 0 6= f ∈ F [x]and c ∈ R. Then f can be uniquely written as f = (x − c)mg, where m ∈ N andg ∈ R[x], g(c) 6= 0. (To see the uniqueness of m and g, note that x − c)h = 0(h ∈ R[x])⇒ h = 0.) m is called the multiplicity of root c of f . c is a multiple rootof f (i.e., with multiplicity m > 1) ⇔ f(c) = f ′(c) = 0.

The Hasse derivative. Let R be a commutative ring. For f(x) = a0 +a1x+· · ·+ anx

n ∈ R[x] and k ≥ 0, define

∂kf =(k

k

)ak +

(k + 1k

)ak+1x+ · · ·+

(n

k

)anx

n−k.

∂kf is called the kth order Hasse derivative of f . We have f (k) = k! ∂kf .

Properties of the Hasse derivative. Let f, g ∈ R[x] and a, b ∈ R.(i) ∂k(af + bg) = a∂kf + b∂kg.(ii) ∂k(fg) =

∑i+j=k(∂if)(∂jg).

(iii) ∂k(f(x+ a)

)= (∂kf)(x+ a).

(iv) For each c ∈ R, f =∑k≥0(∂kf)(c)(x− c)k. In particular, c is a root of f

of multiplicity ≥ m⇔ (∂0f)(c) = · · · = (∂m−1f)(c) = 0.

Definition 2.22 (Content). Let D be a UFD and 0 6= f = a0 + · · ·+ anxn ∈

D[x]. The content of f is C(f) = gcd(a0, . . . , an). If C(f) ∼ 1, f is called primitive.

Lemma 2.23 (Gauss). Let D be a UFD and f, g ∈ D[x] primitive. The fg isprimitive.

Proof. Assume to the contrary that ∃ irreducible p ∈ D such that p | C(fg).Let φ : D[x] → (D/(p))[x] be the homomorphism induced by the natural homo-morphism D → D/(p). Then 0 = φ(fg) = φ(f)φ(g), where φ(f) 6= 0, φ(g) 6= 0.Since D/(p) is an integral domain, so is (D/(p))[x]. We have a contradiction. �

Corollary 2.24. Let D be a UFD and f, g ∈ D[x] nonzero. Then C(fg) ∼C(f)C(g).

Proposition 2.25. Let D be a UFD and F its fractional field. Let f ∈ D[x]be primitive. Then f is irreducible in D[x]⇔ f is irreducible F [x].

Proof. (⇒) Assume to the contrary that f = gh, g, h ∈ F [x], deg g > 0,deg h > 0. Choose a, b ∈ D \ {0} such that ag, bh ∈ D[x]. Then abf = (ah)(bg) ∈D[x]; hence, ab = C(abf) = C(ah)C(bg). So, f = 1

ab (ag)(bh) = agC(ag) ·

bhC(bh) , where

agC(ag) ,

bhC(bh) ∈ D[x] have degree > 0. Contradiction.

(⇐) Obvious. �

2.5. POLYNOMIAL RINGS 45

Theorem 2.26. Let D be a UFD. Then D[x] is also a UFD. The irreducibleelements of D[x] are precisely irreducible elements of D and primitive polynomialsin D[x] which are irreducible in F [x], where F is the fractional field of D.

Proof. The second claim follows from Proposition 2.25. It remains to showthat D[x] is a UFD.

1◦ Existence of factorization.Let f ∈ D[x] be nonzero and nonunit. Since F [x] is a UFD, f = f1 · · · fn,

where fi ∈ F [x] is irreducible. Choose 0 6= ai ∈ D such that aifi ∈ D[x]. Writeaifi = cigi, where ci ∈ D and gi ∈ D[x] is primitive. Then

a1 · · · anf = (a1f1) · · · (anfn) = c1 · · · cng1 · · · gn.

Compare the contents of both sides. We have c1,···cn

a1···an∈ D. Thus,

f =c1, · · · cna1 · · · an

g1 · · · gn,

where c1,···cn

a1···anis a product of irreducibles in D.

2◦ Uniqueness of factorization.Suppose

(2.1) a1 · · · amf1 · · · fn = b1 · · · bsg1 · · · gt,

where a1, . . . , am, b1, . . . , bs ∈ D are irreducible and f1, . . . , fn, g1, . . . , gt ∈ D[x] areirreducible of degree > 0. Compare the contents of the two sides of (2.1). We havea1 · · · am ∼ b1 · · · bs. So, m = s and after reordering, ai ∼ bi.

In F [x],f1 · · · fn ∼ g1 · · · gt.

Thus, n = t and after reordering, fj ∼ gj in F [x]. So, fj = uv gj for some u, v ∈

D \ {0}, i.e., vfj = ugj . Then v = C(ufj) ∼ C(ugj) = u in D. Thus, fj ∼ gj inD[x]. �

Corollary 2.27. If D is a UFD and X is a set of indeterminates, then D[X]is a UFD.

Eisenstein’s criterion. Let D be a UFD with fractional field F and letf = a0 + · · · + anx

n ∈ D[x], n > 0. If there is an irreducible element p ∈ D suchthat p - an, p | ai for 0 ≤ i ≤ n− 1 and p2 - a0, then f is irreducible in F [x].

Proof. Write f = C(f)f1, where f1 ∈ D[x] is primitive. Then the conditionsfor the coefficients of f are also satisfied for the coefficients of f1. So, we may assumethat f is primitive. It suffices to show that f is irreducible in D[x]. Assume to thecontrary that f = gh, g, h ∈ D[x], deg g > 0, deg h > 0. Let φ : D[x]→ (D/(p))[x]be the homomorphism induced by the natural homomorphism D → D/(p). Thenφ(an)xn = φ(g)φ(h). Since D/(p) is an integral domain, we have φ(g) = αxk,φ(h) = βxl, α, β ∈ D/(p). Since k ≤ deg g, l ≤ deg h, but k+ l = n = deg g+deg h,we have k = deg g and l = deg h; hence k, l > 0. Then p | g(0), p | h(0), ⇒ p2 |g(0)h(0) = a0, which is a contradiction. �

Example. Let p be a prime. Then Φp(x) = 1 + x + · · · + xp−1 ∈ Q[x]is irreducible. (Apply Eisenstein’s criterion to Φp(x + 1) = 1

x

[(x + 1)p − 1

]=∑p

i=1

(pi

)xi−1.)

46 2. RINGS AND MODULES

2.6. Modules, Definitions and Basic Facts

Definition 2.28. Let R be a ring (not required to have identity). A left R-modules is an abelian group (A,+) equipped with a scalar multiplication R×A→A, (r, a) 7→ ra such that for r, s ∈ R and a, b ∈ A,

(i) r(a+ b) = ra+ rb;(ii) (r + s)a = ra+ sa;(iii) r(sa) = (as)a.

A right R-module is an abelian group (A,+) equipped with a scalar multiplicationA × R → A. (a, r) 7→ ar such that the analogies of (i) – (iii) hold. A left (right)R-module is sometimes denoted by RA (AR). If R has identity and

(iv) 1Ra = a for all a ∈ A,A is called a unitary left R-module.

Declaration. Unless specified otherwise, all modules are assumed to be uni-tary. A module is assumed to be left if the side is not specified.

Examples of modules. Abelian groups are Z-modules. A vector space overa field F is an F -module. A ring R is an R-module; submodules of RR are leftideals.

Let V be a vector space over a field F and α ∈ HomF (V, V ). For each f ∈ F [x]and v ∈ V , define fv = f(α)v. Then V is an F [x]-module.

Let A be an abelian group. For each a ∈ A and f ∈ End(A), define fa = f(a).Then A is an End(A)-module.

Homomorphism. Let A,B be R-modules. A function f : A → B is called ahomomorphism, or an R-map, if f(a + b) = f(a) + f(b) and f(ra) = rf(a) for alla, b ∈ A and r ∈ R.

Submodule. Let A be an R-module and B ⊂ A. B is called a submodule ofA if B (with the inherited operations) is an R-module.

If X ⊂ A, the smallest submodules of A containing X, called the submodulesgenerated by X, is

〈X〉 ={ n∑i=1

rixi : n ∈ N, ri ∈ R, xi ∈ X}.

Quotient module. Let A be an R-modules and B a submodules of A. LetA/B be the quotient abelian group. For a+B ∈ A/B and r ∈ R, define r(a+B) =ra+B. Then A/B is an R-module, called the quotient module of A by B.

Isomorphism Theorems.

First isomorphism theorem. Let f : A → B be a homomorphism of R-modules. The

f : A/ ker f −→ im f

a+ ker f 7−→ f(a)is an isomorphism.

Second isomorphism theorem. Let A,B be submodules of an R-modules.Then (A+B)/B ∼= A/A ∩B.

Third isomorphism theorem. Let C ⊂ B ⊂ A be R-modules. Then(A/C)/(B/C) ∼= A/B.

2.6. MODULES, DEFINITIONS AND BASIC FACTS 47

Direct product and external direct sum. Let {Ai : i ∈ I} be a familyof R-modules. The direct product of {Ai : i ∈ I}, denoted by

∏i∈I Ai, is the

cartesian product of Ai, i ∈ I. Elements in∏i∈I Ai are of the form (ai)i∈I , where

ai ∈ Ai.∏i∈I Ai is an R-module with addition and scalar multiplication defined

component wise.The external direct sum of {Ai : i ∈ I} is⊕(ex)

i∈IAi =

{(ai)i∈I ∈

∏i∈I

Ai : only finitely many ai 6= 0},

which is a submodule of∏i∈I Ai. If |I| <∞,

⊕(ex)i∈I Ai =

∏i∈I Ai.

Internal direct sum. If {Ai : i ∈ I} is a family of submodules of an R-modules A, the submodule⟨⋃

i∈IAi⟩

={∑i∈I

ai : ai ∈ Ai, ai = 0 for almost all i}

is called the sum of {Ai : i ∈ I} and is denoted by∑i∈I Ai. If Ai ∩

∑j∈I\{i}Aj =

{0} for all i ∈ I, then∑i∈I Ai is called an internal direct sum and is denoted by⊕(in)

i∈I Ai. Moreover, ⊕(ex)i∈I Ai −→

⊕(in)i∈I Ai

(ai)i∈I 7−→∑i∈I ai

is an isomorphism. Most of the time, we write both⊕(ex) and

⊕(in) as⊕

.

Hom. Let RA, RB be R-modules. HomR(RA, RB) = the abelian group of allR-maps from A to B. Let S be anther ring.

(i) If RAS is a bimodule, HomR(RAS , RB) is a left S-module. (For f ∈HomR(RAS , RB), s ∈ S and a ∈ A, define (sf)(a) = f(as).)

(ii) If RBS is a bimodule, HomR(RA, RBS) is a right S-module. (For f ∈HomR(RA, RBS), s ∈ S and a ∈ A, define (fs)(a) = (f(a))s.)

Free modules. Let A be an R-module. A subset X ∈ A is called linearlyindependent if r1x1 + · · · + rnxn = 0 (ri ∈ R, x1, . . . , xn ∈ X distinct) ⇒ r1 =· · · = rn = 0. X is called a basis of A if X is independent and 〈X〉 = A. If A has abasis, A is called a free module (on X); in this case,

A =⊕(in)

x∈XRx ∼=

⊕(ex)

x∈XR.

If all bases of A have the same cardinality, this common cardinality is denoted byrankA. Every R-module is a quotient of a free R-module.

Example (A direct product that is not free).∏∞i=1 Z is not a free

Z-modules. Let

A ={(a1, a2, . . . ) ∈

∞∏i=1

Z : for every k > 0, 2k | ai for almost all i}.

We claim that A is not free. (By Theorem 2.35,∏∞i=1 Z is not free.) Clearly,

|A| ≥ 2ℵ0 > ℵ0. Assume to the contrary that A is free. Then rankA > ℵ0. Everycoset of A in 2A contains an element in

⊕∞i=1 Z. Hence A/2A is countable. So,

dimZ2(A/2A) ≤ ℵ0. However, rankA = dimZ2(A/2A). We have a contradiction.

48 2. RINGS AND MODULES

Theorem 2.29. Let D be a division ring. Then every D-module V is free.Any two bases of V have the same cardinality. V is called a vector space over D;dimD V := |X|, where X is any basis of V .

Proof. A maximal linearly independent subset of V , which exists by Zorn’slemma, is a basis.

Let X and Y be two bases of V . If |X| = ∞ or |Y | = ∞, we have |X| = |Y |by the next lemma. So assume X = {x1, . . . , xn} and Y = {y1, . . . , ym}. Assumeto the contrary that n > m. We have

x1

...xn

= A

y1...ym

,y1...ym

= B

x1

...xn

for some matrices A ∈ Mn×m(D) and B ∈ Mm×n(D). It follows that AB =In. There exists an invertible C ∈ Mn(D) such that CA = [ ∗

0 ··· 0 ]. Thus,(0, . . . , 0, 1)C = (0, . . . , 0, 1)CAB = 0, →←. �

Lemma 2.30. Let R be a ring and F a free R-module with an infinite basis X.Then every basis of F has the same cardinality as X.

Proof. Let Y be another basis of F . We claim that |Y | = ∞. (Otherwise,since each y ∈ Y is a linear combination of finitely many x ∈ X, F is generated bya finite subset X1 of X. But any x ∈ X \X1 is not a linear combination of elementsin X1, →←.)

For each x ∈ X, ∃ a finite subset {y1, . . . , yn} ⊂ Y such that x = r1y1 +· · · + rnyn, ri ∈ R. Define f(x) = {y1, . . . , yn}. We claim that

⋃x∈X f(x) = Y .

(Otherwise, X is spanned by Y1 :=⋃x∈X f(x) ( Y ; hence Y is spanned by Y1,

→←.) Now,

|Y | =∣∣∣ ⋃x∈X

f(x)∣∣∣ ≤ |X|ℵ0 = |X|.

By symmetry, |X| ≤ |Y |. Hence, |X| = |Y |. �

Facts. Let D be a division ring.(i) If W ⊂ V are vector spaces over D, then dimV = dimW + dim(V/W ).(ii) (The dimension formula) If V and W are subspaces of some vector space

over D, then

dimV + dimW = dim(V +W ) + dim(V ∩W ).

Proof. (i) Let X be a basis of W . Extend X to a basis X�∪ Y of V . Then

y +W (y ∈ Y ) are all distinct and form a basis of V/W . So, dimV/W = |Y |.(ii) Define a D-map

f : V ×W −→ V +W

(v, w) 7−→ v + w.

Then f is onto and ker f = {(v,−v) : v ∈ V ∩W} ∼= V ∩W . Hence

dimV + dimW = dim(V ×W ) = dim(V +W ) + dimV ∩W.

�

2.7. PROJECTIVE AND INJECTIVE MODULES 49

The invariant dimension property. A ring R is said to have the invariantdimension property (IDP) if for every free R-module F , any two bases of F havethe same cardinality.

Division rings (Theorem 2.29) and commutative rings (the next theorem) haveIDP. Let A =

⊕∞j=0 Z and R = End(A), then R does not have IDP. For any positive

integer n, partition N as N1 ∪ · · · ∪Nn such that |Ni| = ℵ0. Let τi : Ni → N be abijection. Define fi ∈ End(A) by setting

fi(ej) =

{eτi(j) if j ∈ Ni,0 if j /∈ Ni,

where ej = (0, . . . , 0,j

1, 0, . . . ). Then f1, . . . , fn is a basis of RR. (Proof. ∀h ∈End(A), let gi ∈ End(A) such that gi(eτi(j)) = h(ej). Then

(∑ni=1 gifi

)(ej) = h(ej)

∀j ∈ N. So, h =∑ni=1 gifi; hence f1, . . . , fn generate RR. If

∑i=1 gifi = 0, where

gi ∈ End(A), then gk(A) =(∑n

i=1 gifi)(〈ej : j ∈ Nk〉) = {0}. So, gk = 0 for all

1 ≤ k ≤ n; hence f1, . . . , fn are linearly independent.)

Proposition 2.31. A commutative ring R has IDP.

Proof. Let F be a free R-module and let X be a basis of F . Let I be amaximal ideal of R. Then F/IF is a vector space over R/I.

1◦ We claim that x+IF , x ∈ X, form a basis of R/I(F/IF ). Assume∑ni=1(ai+

I)(xi + IF ) = 0, where ai ∈ F , xi ∈ X (xi distinct). Then∑ni=1 aixi ∈ IF . Hence∑n

i=1 aixi =∑mj=1 bjyj , bj ∈ I, yj ∈ X. It follows that ai ∈ I, 1 ≤ i ≤ n.

2◦ By 1◦, |X| = |{x+ IF : x ∈ X}| = dimR/I(F/IF ), where dimR/I(F/IF ) isindependent of X. �

2.7. Projective and Injective Modules

Exact sequences. A sequence of R-modules and R-maps

· · · −→ Ai−1fi−1−→ Ai

fi−→ Ai+1fi+1−→ · · ·

is called exact if im fi−1 = ker fi for all i. An exact sequence 0→ Af→ B

g→ C → 0

is called a short exact sequence. Two short exact sequences 0→ Af→ B

g→ C → 0

and 0 → A′f ′→ B′ g

′

→ C ′ → 0 are called isomorphic if ∃ isomorphisms α, β, γ suchthat

0 −→ Af−→ B

g−→ C −→ 0

α

y β

y γ

y0 −→ A′

f ′−→ B′ g′−→ C ′ −→ 0commutes.

Proposition 2.32. Let 0 → Af→ B

g→ C → 0 be a short exact sequence ofR-modules. Then the following statements are equivalent.

(i) ∃ an R-map h : C → B such that g ◦ h = idC .(ii) ∃ an R-map k : B → A such that k ◦ f = idA.

(iii) 0→ Af→ B

g→ C → 0 is isomorphic to 0→ Aι1→ A⊕ C π2→ C → 0.

If (i) – (iii) are satisfied, the short exact sequence 0 → Af→ B

g→ C → 0 is calledsplit.

50 2. RINGS AND MODULES

Proof. (i) ⇒ (iii).

................................................................................................. ........... ........................................................................................ ........... ........................................................................................ ........... ................................................................................................. ...........

................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ....................................

................

................

................

................

................

............................

................

................

................

................

................

............................

................

................

................

................

................

............................................................................

0

0

A

A

A⊕ C

B

C

C

0

0f g

h

ι1 π2

idA φ idC

commutes, whereφ : A⊕ C −→ B

(a, c) 7−→ f(a) + h(c)

is an isomorphism by the five lemma (next).(ii) ⇒ (iii).

................................................................................................. ........... ........................................................................................ ........... ........................................................................................ ........... ................................................................................................. ...........

................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ................................................................................................................................................

............................................................................................................

............................................................................................................

................................................

0

0

A

A

A⊕ C

B

C

C

0

0f g

k

ι1 π2

idA ψ idC

commutes, whereψ : B −→ A⊕ C

b 7−→(k(b), g(b)

)is an isomorphism by the five lemma.

(iii) ⇒ (i) and (ii).

................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ...........

................................................................................................. ........... ..................................................................................... ........... ..................................................................................... ........... ................................................................................................. ................................................................................................................................................

............................................................................................................

............................................................................................................

.........................

......................... .........................

........................................

................................................

........................................

................................................0

0

A

A A⊕ C

B C

C

0

0

f g

k h

ι1 π2

α β γ

ι2π1

Let k = α ◦ π1 ◦ β−1, h = β ◦ ι2 ◦ γ−1. �

Lemma 2.33 (The five lemma). Let

A1f1−→ A2

f2−→ A3f3−→ A4

f4−→ A5

α1

y α2

y α3

y α4

y α5

yB1

g1−→ B2g2−→ B3

g3−→ B4g4−→ B5

be a commutative diagram of R-modules with exact rows.(i) If α1 is surjective and α2, α4 are injective, then α3 is injective.(ii) If α5 is injective and α2, α4 are surjective, then α3 is surjective.

Proof. (i) Let a3 ∈ kerα3. Then α4f3(a3) = g3α3(a3) = 0. Since α4 isinjective, f3(a3) = 0. So, a3 = f2(a2) for some a2 ∈ A2. Let b2 = α2(a2). Theng2(b2) = α3(a3) = 0. So, b2 = g1(b1) for some b1 ∈ B1. Let a1 ∈ A1 such thatα1(a1) = b1. Then α2(a2−f1(a1)) = α2(a2)−α2f1(a1) = b2−g1α1(a1) = b2−b2 =0. So, a2 = f1(a1). Hence, a3 = f2(a2) = 0.

2.7. PROJECTIVE AND INJECTIVE MODULES 51

(ii) Let b3 ∈ B3. Then g3(b3) = α4(a4) for some a4 ∈ A4. Since α5f4(a4) =g4α4(a4) = g4g3(b3) = 0, we have f4(a4) = 0. So, a4 = f3(a3) for some a3 ∈ A3.Since g3(b3−α3(a3)) = α4(a4)−g3α3(a3) = α4(a4)−α4f3(a3) = α4(a4)−α4(a4) =0, b3−α3(a3) = g2(b2) for some b2 ∈ B2. Let a2 ∈ B2 such that b2 = α2(a2). Thenα3(a3+f2(a2)) = α3(a3)+α3f2(a2) = α3(a3)+g2α2(a2) = α3(a3)+g2(b2) = b3. �

Projective modules. An R-module P is called projective if for every sur-jection p : A → B and homomorphism f : P → B, there exists a homomorphismg : P → A such that

............................................................................ ........... ............................................................................ ...........

.................................................................................................................

.................................................0A B

P

p

gf

commutes.Free modules are projective.

Theorem 2.34 (Characterizations of projective modules). Let P be an R-module. The following statements are equivalent.

(i) P is projective.(ii) Every short exact sequence 0→ A

i→ Bp→ P → 0 is split.

(iii) There exists an R-module K such that K ⊕ P is free.

Proof. (i) ⇒ (ii).

............................................................................ ........... ............................................................................ ........... ............................................................................ ........... ............................................................................ ...........

.................................................................................................................

.................................................0 A B P 0

P

i

g

p

id

(ii) ⇒ (iii). There exists a free R-module F and surjection p : F → P . Since0→ ker p ↪→ F

p→ P → 0 is exact, hence split, F ∼= ker p⊕ P .(iii) ⇒ (i).

............................................................................ ........... ............................................................................ ...........

.......................................................................................

..............................................................................................................................................................................

.....................................................

.................................................

........................................................................................................

F = K ⊕ P

A B 0

P

π ι

g1

g

p

f

F is projective. So, there exists g1 : F → A such that the diagram commutes. Letg = g1 ◦ ι. �

Pull back. Let

(2.2)

AyfB −→

gC

52 2. RINGS AND MODULES

be a diagram of R-modules. Define D = {(a, b) ∈ A × B : f(a) = g(b)} andα : D → A, (a, b) 7→ a; β : D → B, (a, b) 7→ b. Then

Dα−→ A

β

y yfB −→

gC

is a commutative diagram of R-modules. (D,α, β) is called the pull back of (2.2).In Theorem 2.34, (ii) ⇒ (i) can also be proved using a pull back:

........................................................................................ ........... ......................................................................................... ........... ................................................................................................. .................................... ................................................................................................. ...........

................................................................................................. ........... ................................................................................................. ......................................

............................................................................................................

............................................................................................................

.......................

.................................................................

................................................kerα

A B

0 D P 0

0

α

β f

p

Note that p is onto ⇒ α is onto.

Example. R = Z6. RZ3 is projective (Z3 ⊕ Z2∼= R) but not free.

Theorem 2.35. Let F be a free module over a PID R and A a submodule ofF . Then A is free with rankA ≤ rankF .

Proof. Let X be a basis of F . Let

Y = {(Y,Z, f) : Z ⊂ Y ⊂ X, f : Z → A ∩ 〈Y 〉 1-1, f(Z) is a basis of A ∩ 〈Y 〉}.For (Y1, Z1, f1), (Y2, Z2, f2) ∈ Y, define (Y1, Z1, f1) ≺ (Y2, Z2, f2) if Y1 ⊂ Y2, Z1 ⊂Z2 and f2|Z1 = f1. Then (Y,≺) is a poset in which every chain has an upperbound. By Zorn’s lemma, (Y,≺) has a maximal element (Y0, Z0, f0). It suffices toshow Y0 = X.

Suppose to the contrary that Y0 6= X. Let x0 ∈ X \ Y0. Put

I = {r ∈ R : rx0 + y ∈ A for some y ∈ 〈Y0〉}.I is an ideal of R; hence I = 〈s〉 for some s ∈ R. If s = 0, A∩〈Y0∪{x0}〉 = A∩〈Y0〉.Then (Y0 ∪ {x0}, Z0, f0) � (Y0, Z0, f0), →←. So, s 6= 0. Let u ∈ A such thatu = sx0 + y for some y ∈ 〈Y0〉. We claim that

A ∩ 〈Y0 ∪ {x0}〉 = A ∩ 〈Y0〉 ⊕ 〈u〉.If w ∈ A ∩ 〈Y0 ∪ {x0}〉, then w = tx0 + z for some z ∈ 〈Y0〉 and t ∈ R with s | t.So, w − t

su ∈ A ∩ 〈Y0〉 ⇒ w ∈ A ∩ 〈Y0〉 + 〈u〉. Now f0(Z0) ∪ {u} is a basis ofA ∩ 〈Y0 ∪ {x0}〉. Extend f0 : Z → A ∩ 〈Y0〉 to g : Z0 ∪ {x0} → A ∩ 〈Y0 ∪ {x0}〉 bysetting g(x0) = u. Then (Y0 ∪ {x0}, Z0 ∪ {x0}, g) � (Y0, Z0, f0). →←. �

Note. If rankF <∞, Theorem 2.35 can be proved by an induction on rankF ;the argument is similar to the above proof but Zorn’s lemma is not needed.

Theorem 2.36. Every projective module over a PID is free.

Proof. Let P be a projective module over a PID R. By Theorem 2.35 (iii),P is a submodule of a free R-module. By Theorem 2.35, P is free. �

Theorem 2.37 (Serre’s conjecture, Quillen-Suslin’s theorem). Let k be a field.Then every projective module over k[x1, . . . , xn] is free.

2.7. PROJECTIVE AND INJECTIVE MODULES 53

Projective modules over a local ring.

Theorem 2.38 (Kaplansky [7]). Every projective module over a local ring (notnecessarily commutative) is free.

Lemma 2.39. If A is a direct sum of countably generated R-modules and B isa direct summand of A, then B is a direct sum of countably generated R-modules.

Proof. Let A =⊕

i∈I Ai, where Ai is countably generated. Let A = B ⊕ C.For each J ⊂ I, put AJ =

∑i∈J Ai. Let

X = {J ⊂ I :AJ = AJ ∩B +AJ ∩ C, AJ ∩B is a direct sum

of countably generated R-modules}.

(X ,⊂) is a poset in which every chain has an upper bound (the union of the setsin the chain). By Zorn’s lemma, (X ,⊂) has a maximal element J0.

We claim that J0 = I. (The conclusion of the lemma follows from the claim.)Assume to the contrary that ∃i1 ∈ I \ J0. Let J1 = {i1} and AJ1 = 〈x11, x12, . . . 〉.Write x1j = x′1j + x′′1j , where x′1j ∈ B, x′′2j ∈ C. Each x′1j (x′′1j) is contained in AJfor some finite J ⊂ I. So,

⋃∞j=1{x′1j , x′′1j} ⊂ AJ2 for some countable J2 ⊂ I. Write

AJ2 = 〈x21, x22, . . . 〉, x2j = x′2j + x′′2j , x′2j ∈ B, x′′2j ∈ C. Then

⋃∞j=1{x′2j , x′′2j} ⊂

AJ3 for some countable J3 ⊂ I. In general,

AJi⊂ AJi+1 ∩B +AJi+1 ∩ C.

Let J∗ =⋃∞i=0 Ji. Then

AJ∗ ⊂ AJ∗ ∩B +AJ∗ ∩ C.

Since AJ0 ∩ B is a direct summand of AJ0 and AJ0 is a direct summand of A,AJ0 ∩B is a direct summand of A. Hence AJ0 ∩B is a direct summand of AJ∗ ∩B.(Cf. Exercise 2.6.) Since AJ∗ = AJ∗ ∩B ⊕AJ∗ ∩C and AJ0 = AJ0 ∩B ⊕AJ0 ∩C,we have

AJ∗

AJ0

=AJ∗ ∩BAJ0 ∩B

⊕ AJ∗ ∩ CAJ0 ∩ C

.

Thus, (AJ∗ ∩ B)/(AJ0 ∩ B) is a homomorphic image of AJ∗/AJ0 . Since AJ∗ iscountably generated, so is (AJ∗ ∩B)/(AJ0 ∩B). Hence

AJ∗ ∩B ∼= (AJ0 ∩B)⊕((AJ∗ ∩B)/(AJ0 ∩B)

)is a direct sum of countably generated R-modules. Thus J∗ ∈ X , which contradictsthe maximality of J0. �

Proof of Theorem 2.38. Let R be a local ring with maximal ideal m. LetP be a projective module over R.

1◦ Every x ∈ P is contained in a free direct summand of P .There exists an R-module Q such that F := P ⊕Q is free. Let U be a basis of

F . Write x = a1u1 + · · · anun, ai ∈ R, u1, . . . , un ∈ U distinct. Assume U is chosensuch that n is as small as possible. Then for each 1 ≤ i ≤ n,

(2.3) ai /∈ a1R+ · · ·+ ai−1R+ ai+1R+ · · ·+ anR.

(If an = a1b1 + · · · an−1bn−1, then x = a1(u1 + b1un) + · · ·+ an−1(un−1 + bn−1un).Note that {u1 + b1un, . . . , un−1 + bn−1un, un} ∪ U ′ is a basis of F , where U ′ =

54 2. RINGS AND MODULES

U \ {u1, . . . , un}. This contradicts the minimality of n.) Write ui = yi + zi, yi ∈ P ,zi ∈ Q. Then

(2.4) a1u1 + · · ·+ anun = a1y1 + · · ·+ anyn.

Write

(2.5)

y1...yn

≡ Cu1

...un

(mod 〈U ′〉).

By (2.4) and (2.5), we have

[a1, . . . , an] = [a1, . . . , an]C,

i.e., [a1, . . . , an](I−C) = 0. By (2.3), all entries of I−C are in m. Since R is local,C is irreducible in Mn×n(R). So, by (2.5), {y1, . . . , yn} ∪ U ′ is a basis of F . LetY = 〈y1, . . . , yn〉. Then x ∈ Y and Y is free and is a direct summand of F hence adirect summand of P .

2◦ P is a direct summand of a free R-module. By Lemma 2.39, P is a directsum of countably generated R-modules. Thus we may assume that P is countablygenerated.

Let P = 〈x1, x2, . . . 〉. By 1◦, P = F1⊕P1, where F1 is free and x1 ∈ F1. Writex2 = x′2 + x′′2 , x′2 ∈ F1, x′′2 ∈ P1. By 1◦ again, P1 = F2 ⊕ P2, where F2 is free andx′′2 ∈ F2. Write x3 = x′3+x′′3 , x′3 ∈ F1⊕F2, x′′3 ∈ P2, ... Then P = F1⊕F2⊕· · · . �

Injective module. An R-module E is called injective if for every injectioni : A→ B and homomorphism f : A→ E, there exists a homomorphism g : B → Esuch that

............................................................................ ........... ............................................................................ .................................................................................................. ..........................

.................................................

0 A B

E

i

gf

commutes.

Fact. Let {Ei : i ∈ I} be a family of R-modules. Then∏i∈I Ei is injective ⇔

Ei is injective for all i ∈ I.

Proof. (⇒)

............................................................................ ........... ............................................................................ ..................................................................................................

..............................................................................................................................................................................

..........................

.....................................................................................................................................................................................................................

.................................................

0 A B

Ei

∏i∈I Ei

i

f

ιi πi

2.7. PROJECTIVE AND INJECTIVE MODULES 55

(⇐)............................................................................ ........... ............................................................................ ...........

.......................................................................................

.......................................................................................

..........................

.....................................................................................................................................................................................................................

.................................................

0 A B

Ei

∏i∈I Ei

i

f

hi

h

πi

h(b) = (hi(b))i∈I .

�

Push out. Let

(2.6)A

f−→ B

g

yC

be a diagram of R-modules. Let S ={(f(a),−g(a)) : a ∈ A

}⊂ B ⊕ C, D =

(B ⊕ C)/S, α : B → D, b 7→ (b, 0) + S, β : C → D, c 7→ (0, c) + S. Then

Af−→ B

g

y yαC −→

βD

is a commutative diagram of R-modules. (D,α, β) is called the push out of (2.6).

Proposition 2.40 (Characterizations of injective modules). Let E be an R-module. The following statements are equivalent.

(i) E is injective.(ii) Every short exact sequence 0→ E

i→ Ap→ B → 0 is split.

(iii) If E is a submodule of A, then A = E ⊕B for some submodule B of A.

Proof. (i) ⇒ (ii).

............................................................................ ........... ............................................................................ ........... ............................................................................ ........... ............................................................................ .................................................................................................. ..........................

.................................................

0 A B 0E

E

i p

gid

(ii) ⇒ (i). Use a push out................................................................................................. ........... ................................................................................................. ...........

................................................................................................. ........... ................................................................................................. .................................... ................................................................................... ........... ................................................................................... ......................................

............................................................................................................

............................................................................................................

.................................................................

................................................

0 A B

0 E D cokerβ 0

i

f

β

α

Note that i is 1-1 ⇒ β is 1-1. (If x ∈ kerβ, (x, 0) ∈ S, i.e., (x, 0) = (f(a),−i(a))for some a ∈ A. So, i(a) = 0⇒ a = 0⇒ x = f(a) = 0.)

(ii) ⇒ (iii). 0→ E ↪→ A→ A/E → 0 is split.(iii) ⇒ (ii). Obvious. �

56 2. RINGS AND MODULES

Note. Theorem 2.44 also provides a quick proof of (iii) ⇒ (i).

Theorem 2.41 (Baer’s criterion). An R-module E is injective⇔ given any leftideal L of R and R-map α : L→ E, α can be extended to an R-map β : R→ E.

Proof. (⇐) Given0 −→ A

i−→ B

f

yE

May assume that A ⊂ B and i is the inclusion. Let

S = {(C, h) : A ⊂ RC ⊂ B, h : C → E is an R-map, h|A = f}.For (C1, h1), (C2, h2) ∈ S, define (C1, h1) ≺ (C2, h2) if C1 ⊂ C2 and h2|C1 = h1.(S,≺) is a nonempty poset in which every chain has an upper bound. By Zorn’slemma, (S,≺) has a maximal element of (C0, h0). It remains to show that C0 = B.

Assume to the contrary that ∃b ∈ B \ C0. Let L = {r ∈ R : rb ∈ C0}. L is aleft ideal of R. α : L → E, r 7→ h0(rb) is an R-map. So, α extends to an R-mapβ : R→ E. Define

h1 : C0 +Rb −→ E

c+ rb 7−→ h0(c) + rβ(1)

h1 is a well-defined R-map. (If c + rb = c′ + r′b, then (r − r′)b = c′ − c ∈ C0. So,h0(c′ − c) = h0((r − r′)b) = α(r − r′) = β(r − r′) = (r − r′)β(1).) Also h1|C0 = h0.So, (C0 +Rb, h1) � (C0, h0), →←. �

Divisible modules. Let R be an integral domain and D and R-module. Dis called divisible if ∀y ∈ D, and 0 6= r ∈ R, ∃x ∈ D such that rx = y.

Facts.

(i) Di, i ∈ I divisible ⇔⊕

i∈I Di divisible.(ii) D divisible and E ⊂ D ⇒ D/E divisible.(iii) D injective ⇒ D divisible.

Proposition 2.42. Let D be a modules over a PID R. Then D is injective ⇔D is divisible.

Proof. (⇐) Let I 6= 0 be an ideal of R and f : I → D an R-map. We haveI = 〈a〉 for some 0 6= a ∈ R. Since D is divisible, ∃x ∈ D such that ax = f(a).Define g : R → D, r 7→ rx. Then g is an R-map and g|I = f . By Baer’s criterion,D is injective. �

Proposition 2.43. Every abelian group A can be embedded in a divisibleabelian group.

Proof. A ∼= (⊕

i∈I Z)/K ↪→ (⊕

i∈I Q)/K, where (⊕

i∈I Q)/K is divisible.�

Theorem 2.44. Every R-module A can be embedded in an injective R-module.

Proof. By Proposition 2.43, ∃Z-module embedding f : A→ B, where B is adivisible abelian group. Then we have R-module embeddings

Aφ−→ HomZ(ZRR, ZA)

f−→ HomZ(ZRR, ZB)

2.8. CHAIN CONDITIONS 57

whereφ(a) : R −→ A

r 7−→ ra

f(α) : R −→ B

r 7−→ f(α(r))By the next lemma, HomZ(ZRR, ZB) is an injective R-modules. �

Lemma 2.45. Let R be a ring and B a divisible abelian group. Then HomZ(ZRR, ZB)is an injective R-module.

Proof. Let L be a left ideal of R and f : L→ HomZ(R,B) an R-map. Let

g : L −→ B

x 7−→ [f(x)](1R).

g is a Z-map. So, g extends to a Z-map g : R→ B. For each r ∈ R, define

h(r) : R −→ B

y 7−→ g(yr).

Then h(r) ∈ HomZ(R,B), h : R → HomZ(R,B) is an R-map and h|L = f . ByBaer’s criterion, HomZ(R,B) is injective. �

2.8. Chain Conditions

Let RA be an R-module. Two descending (or ascending) sequences of submod-ules

A = A0 ⊃ A1 ⊃ · · · ⊃ An = {0}A = A′0 ⊃ A′1 ⊃ · · · ⊃ A′m = {0}

are called equivalent if n = m and ∃ a permutation σ of {1, . . . , n} such thatAi−1/Ai ∼= A′σ(i)−1/A

′σ(i). A descending sequence A = A0 ⊃ A1 ⊃ · · · ⊃ An = {0}

is called a composition series of A if Ai−1/Ai is simple for all 1 ≤ i ≤ n.

Theorem 2.46 (Jordan-Holder). Any two composition series of a module RAare equivalent.

Proof. Same as the proof of the Jordan-Holder theorem for groups. �

ACC and DCC. An R-module A is said to have the ascending chain condition(ACC) if for every ascending chain of submodules A1 ⊂ A2 ⊂ · · · , there exists nsuch that An = An+1 = · · · . A is said to have the descending chain condition(DCC) if for every descending chain of submodules A1 ⊃ A2 ⊃ · · · , there exists nsuch that An = An+1 = · · · .

Proposition 2.47. A module RA has a composition series ⇔ A has both ACCand DCC.

Proof. (⇐) We construct a composition series A = A0 ⊃ A1 ⊃ · · · as follows.Let A0 = A. If A0 6= 0, since A has ACC, among all proper submodules of A0,there is a maximal one, say, A1. Clearly, A0/A1 is simple. By induction, there aresubmodules A0 ⊃ A1 ⊃ A2 ⊃ · · · such that Ai/Ai+1 is simple for all i and Ai+1 isdefined whenever Ai 6= 0. Since A has DCC, the above descending series must stopat An. So, An = 0. Now, A = A0 ⊃ A1 ⊃ · · · ⊃ An = 0 is a composition series ofA. �

Proposition 2.48. Let A be an R-module.

58 2. RINGS AND MODULES

(i) A has ACC ⇔ every nonempty family of submodules of A contains amaximal element ⇔ every submodule of A is finitely generated.

(ii) A has DCC ⇔ every nonempty family of submodules of A contains aminimal element.

Definition 2.49. A ring R is called left (right) noetherian if the module RR(RR) has ACC. R is called left (right) artinian if the module RR (RR) has DCC.R is called noetherian (artinian) if it is both left and right noetherian (artinian).

The Hopkins-Levitzki theorem (Theorem ??). A left (right) artinianring is left (right) noetherian.

Proof. Not easy, will be given in §??. �

Theorem 2.50 (Hilbert basis theorem). If R is a left (right) commutativenoetherian ring, then so is R[x1, . . . , xn].

Proof. We only have to show that R[x] is left noetherian. Assume to thecontrary that there exists a left ideal I of R[x] which is not finitely generated. Letf0 ∈ I be a polynomial of the smallest degree. Then I 6= (f0). Let f1 ∈ I \ (f0)be of the smallest degree. In general, let fn+1 ∈ I \ (f0, . . . , fn) be of the smallestdegree. Let dn = deg fn. Then d0 ≤ d1 ≤ · · · . Let an be the leading coefficient offn. Then (a0) ⊂ (a0, a1) ⊂ · · · is an ACC of RR. Since R is left noetherian, ∃msuch that (a0, . . . , am) = (a0, . . . , am, am+1). So,

am+1 = r0a0 + · · ·+ rmam, ri ∈ R.Put

f = fm+1 −m∑i=0

rifi(x)xdm+1−di .

Then f ∈ I \ (f0, . . . , fm) and deg f < dm+1, which is a contradiction. �

Proposition 2.51. Let 0 → Ai→ B

p→ C → 0 be an exact sequence of R-modules. Then B has ACC (DCC) ⇔ both A and C have ACC (DCC).

Proof. B has ACC ⇒ A and C have ACC.Let A1 ⊂ A2 ⊂ · · · be an ascending sequence of submodules of A. Then

i(A1) ⊂ i(A2) ⊂ · · · is an ascending sequence of submodules of B. Thus i(A1) ⊂i(A2) ⊂ · · · stabilizes and so does A1 ⊂ A2 ⊂ · · · .

Let C1 ⊂ C2 ⊂ · · · be an ascending sequence of submodules of C. Thenp−1(C1) ⊂ p−1(C2) ⊂ · · · is an ascending sequence of submodules of B, so itstabilizes. Since Ci = p(p−1(Ci)), C1 ⊂ C2 ⊂ · · · also stabilizes.

A and C have ACC ⇒ B has ACC.Let B1 ⊂ B2 ⊂ · · · be an ascending sequence of submodules of B. Then ∃n > 0

such that for all k > 0, p(Bn) = p(Bn+k) and i−1(Bn) = i−1(Bn+k). We have acommutative diagram

................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ...........

................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ........... ................................................................................................. ...........

............................................................................................................

............................................................................................................

.....................................................................................................

.......................0 i−1(Bn) Bn p(Bn) 0

0 i−1(Bn+k) Bn+k p(Bn+k) 0

i p

i p

id id

By the five lemma, Bk = Bn+k. �

2.9. FINITELY GENERATED MODULES OVER A PID 59

Proposition 2.52. Let R be a left noetherian (artinian) ring. Then everyfinitely generated R-module A has ACC (DCC).

Proof. A ∼= Rn/K. Since R has ACC, by Proposition 2.51, Rn and Rn/Khas ACC. �

Proposition 2.53. Let 0 → Ai→ B

p→ C → 0 be an exact sequence of R-modules.

(i) Assume that A = 〈X〉 and C = 〈Y 〉. Choose Z ⊂ B such that p(Z) = Y .Then B = 〈X ∪Z〉. In particular, A and C are finitely generated ⇒ B isfinitely generated.

(ii) If R is left noetherian, then B is finitely generated ⇔ both A and C arefinitely generated.

Proof. (⇒) By Proposition 2.54 (i), A is finitely generated. �

Proposition 2.54. Let R be a left noetherian ring and M a finitely generatedR-module.

(i) Every submodule of M is finitely generated.(ii) If R is a PID and M is generated by n elements, then every submodules

of M can be generated by ≤ n elements.

Proof. (i) Let M = 〈x1, . . . , xn〉 and let S be a submodule of M . Usinginduction on M .

If n = 1, M = 〈x1〉 ∼= R/I for some left ideal I of R. Then S ∼= J/I for aomeleft ideal J of R with J ⊃ I. Since R is left noetherian, J is fnitely generated andso is J/I.

Assume n > 1. Let M1 = 〈x1, . . . , xn−1〉. Then

0→ S ∩M1 → S → S/(S ∩M1)→ 0

is exact. Since S ∩ M1 ⊂ M1, by the induction hypothesis, S ∩ M1 is finitelygenerated. Since S/(S ∩M1) ∼= (S +M1)/M1 ⊂ M/M1 = 〈xn +M1〉, S/(S ∩M1)is also finitely generated. Thus S is finitely generated.

(ii) In the proof of (i), S/(S ∩M1) is cyclic. �

2.9. Finitely Generated Modules over a PID

Theorem 2.55 (Structure of finitely generated modules over a PID). Let A bea finitely generated module over a PID R. Then

(2.7) A = Rz1 ⊕ · · · ⊕Rzs,

where

(2.8) R 6= ann(z1) ⊃ · · · ⊃ ann(zs).

Moreover, ann(z1), · · · , ann(zs) are uniquely determined by (2.7) and (2.8). (Note.Rzi ∼= R/ann(zi).)

Proof. Existence of decomposition (2.7).Since A is finitely generated, we may assume A = Rn/K, where K is a sub-

module of Rn. Since R is a PID, by Theorem 2.35, K is free of rank m ≤ n. (We

60 2. RINGS AND MODULES

only need that fact K is finitely generated.) Let K = (f1, . . . , fm) and writef1...fm

= A

e1...en

,where e1, . . . , en is the standard basis of Rn and A ∈ Mm×n(R). There existP ∈ GL(m,R) and Q ∈ GL(n,R) such that

PAQ =

d1

. . . 0dr

0 0

,where di 6= 0, d1 | d2 | · · · | dr. (This is the Smith normal form of A.) We assumed1 = · · · = da = 1 and da+1 /∈ R×. Let

P

f1...fm

=

f ′1...f ′m

and Q−1

e1...en

=

e′1...e′n

.Then

f ′1...f ′m

=

d1

. . . 0dr

0 0

e′1...e′n

.So, K = (f ′1, . . . , f

′m) = (d1e

′1, . . . , dre

′r). Since

Rn = Re′1 ⊕ · · · ⊕Re′n,K = Rd1e

′1 ⊕ · · · ⊕Rdne′n (di = 0 for i > r),

we haveA = Rn/K ∼= Re′1/Rd1e

′1 ⊕ · · · ⊕Re′n/Rdne′n

∼= R/(d1)⊕ · · · ⊕R/(dn)∼= R/(da+1)⊕ · · · ⊕R/(dn).

Let wi = 1+(di) ∈ R/(di), a+1 ≤ i ≤ n. Then R/(di) = Rwi, ann(wi) = (di) and

A ∼= Rwa+1 ⊕ · · · ⊕Rwn.

Uniqueness of ann(z1), . . . , ann(zs).Assume that

A = Rz1 ⊕ · · · ⊕Rzs = Rw1 ⊕ · · · ⊕Rwt,

where R 6= ann(z1) ⊃ · · · ⊃ ann(zs) and R 6= ann(w1) ⊃ · · · ⊃ ann(wt). We willshow that s = t and ann(zi) = ann(wi).

Without loss of generality, assume s ≥ t. Let (w′1, . . . , w′s) = (0, . . . , w1, . . . , wt).

Then

(2.9) A = Rz1 ⊕ · · · ⊕Rzs = Rw′1 ⊕ · · · ⊕Rw′s,

2.9. FINITELY GENERATED MODULES OVER A PID 61

where ann(z1) ⊃ · · · ⊃ ann(zs) and ann(w′1) ⊃ · · · ⊃ ann(w′s). It suffices to showthat ann(zi) = ann(w′i) for all 1 ≤ i ≤ s.

First, ann(zs) = annA = ann(w′s). Let 1 ≤ i < s and let ann(zi) = (di). By(2.9),

Rdizi+1 ⊕ · · · ⊕Rdizs ⊃ Rdiw′i ⊕ · · · ⊕Rdiw′s.

So,

di

w′i...w′s

= diA

zi+1

...zs

, A ∈M(s−i+1)×(s−i)(R).

There exists P ∈ GL(s− i+ 1, R) such that PA = [ ∗0 ··· 0 ]. Hence,

diP

w′i...w′s

= diPA

zi+1

...zs

=

∗...∗0

.Write P = [ ∗

pi ··· ps ]. Then

di[pi, . . . , ps]

w′i...w′s

= 0.

So, dipjw′j = 0, i ≤ j ≤ s, since Rw′i ⊕ · · · ⊕ Rw′s is a direct sum. So, dipj ∈ann(w′j) ⊂ ann(w′i), i ≤ j ≤ s. Since P is invertible, gcd(pi, . . . , ps) = 1. Thus,di ∈ ann(w′i). So, ann(zi) = (di) ⊂ ann(w′i). By symmetry, ann(w′i) ⊂ ann(zi). �

Note. In the above theorem, assume ann(zi) = (di), 1 ≤ i ≤ s, dt 6= 0,dt+1 = · · · = ds = 0. Write

di = pei11 · · · p

eik

k , 1 ≤ i ≤ t,

where p1, . . . , pk ∈ R are distinct irreducibles and eij ∈ N. Then

A ∼= R/(d1)⊕ · · · ⊕R/(dt)⊕Rs−t ∼=[ ⊕

1≤i≤t1≤j≤k

R/(peij

j )]⊕Rs−t.

The integer s − t is called the rank of A; d1, . . . , dt are called the invariant factorof A; peij

j with eij > 0 are called the elementary divisors of A.Two finitely generated modules over a PID are isomorphic iff they have the

same rank and the same invariant factors (elementary divisors).

Example. Let

A =

−18 7 91 −14 8714 −5 3 10 78 −3 3 6 5

126 −47 −275 94 −243

62 2. RINGS AND MODULES

and A = Z/{Ax : x ∈ Z}. The Smith normal form of A is1

220

0 0

.So, A ∼= Z2 ⊕ Z20 ⊕ Z2. The elementary divisors of A are 2, 22, 5; rankA = 2.

The structure theorem of finitely generated modules over a PID can also bederived by the following method. The advantage of the above method is that itallows one to compute the invariant factors.

Another proof of Theorem 2.55. Let A be a finitely generated moduleover a PID R.

Existence of the decomposition of A.1◦ Let Ator = {a ∈ A : ra = 0 for some 0 6= r ∈ R}. Then A/Ator is torsion

free. By the next lemma, A/Ator is a free R-module. Thus the exact sequence0→ Ator ↪→ A→ A/Ator → 0 is split. So,

A ∼= Ator ⊕ (A/Ator).

2◦ For each irreducible p ∈ R, let

A(p) = {a ∈ A : pna = 0 for some n > 0}.Then

Ator =⊕p

A(p),

where the sum is over finitely many irreducibles p ∈ R.3◦ Assume pnA(p) = 0 but pn−1A(p) 6= 0. Let a ∈ A(p) such that pn−1a 6= 0.

Then Ra ∼= R/(pn) (as R-modules and as R/(pn)-modules). Using Baer’s criterion,it is easy to see that R/(pn) is an injective R/(pn)-module. Since Ra is an injectivesubmodule of A(p) (as R/(pn)-modules), we have A(p) = Ra⊕B for some R/(pn)-and R-submodule B of A(p). Apply the same argument to B. ... Since A(p) isfinitely generated, it has ACC (Proposition 2.52). So eventually,

A(p) ∼= R/(pn1)⊕ · · · ⊕R/(pnk).

Uniqueness of the decomposition of A. Let

A = Rr ⊕[⊕p

(R/(pn(p,1))⊕ · · · ⊕R/(pn(p,ip))

)].

Then r = rank(A/Ator) and

dimA/(p)

(pn−1A/pnA

)=∣∣{1 ≤ i ≤ ip : n(p, i) ≥ n}

∣∣.�

Lemma 2.56. Let R be a PID. If A is a finitely generated torsion free R-module,then A is free.

Proof. Assume A = 〈x1, . . . , xn〉. Let {y1, . . . , ym} be a maximal linearlyindependent subset of {x1, . . . , xn}. Then for every 1 ≤ i ≤ n, ∃ 0 6= ai ∈ R suchthat aixi ∈ 〈y1, . . . , ym〉. Let a = a1 · · · an. Then aA ⊂ 〈y1, . . . , ym〉 ∼= Rm. So, aAis free. Since A is torsion free, aA ∼= A. �

2.9. FINITELY GENERATED MODULES OVER A PID 63

The rational canonical form of a linear transformation. Let V bean n-dimensional vector space over F with a basis ε1, . . . , εn. Let T ∈ EndF (V )such that

T

ε1...εn

= A

ε1...εn

, A ∈Mn(F ).

For each f ∈ F [x] and v ∈ V , define fv = f(T )v. Then V is an F [x]-module.Define

φ : F [x]n −→ V

(f1, . . . , fn) 7−→ (f1, . . . , fn)

ε1...εn

.Then φ is an F [x]-map with

(2.10) kerφ = {y(xI −A) : y ∈ F [x]n}.

Proof of (2.10): ∀(f1, . . . , fn) ∈ F [x]n, by the division algorithm, (f1, . . . , fn) =y(xI −A) + (a1, . . . , an) for some y ∈ F [x]n and (a1, . . . , an) ∈ Fn. Then

(f1, . . . , fn)

ε1...εn

=(y(xI −A) + (a1, . . . , an)

)ε1...εn

= (a1, . . . , an)

ε1...εn

.Hence (f1, . . . , fn) ∈ kerφ⇔ (a1, . . . , an) = 0.

Therefore, we have an F [x]-module isomorphism

V ∼= F [x]n/{y(xI −A) : y ∈ F [x]n} = F [x]n/(α1, . . . , αn),

where

xI −A =

α1

...αn

and (α1, . . . , αn) is the F [x]-module generated by α1, . . . , αn. Let the Smith normalform of xI −A be

1. . .

1d1

. . .

dr

.

Then by the proof of Theorem ??,

V ∼= F [x]/(d1)⊕ · · · ⊕ F [x]/(dr),

64 2. RINGS AND MODULES

i.e., V = V1⊕· · ·⊕Vr, where Vi ∼= F [x]/(di). Let di = xei +ai,ei−1xei−1 + · · ·+ai,0.

Then 1, x, . . . , xei−1 is an F -basis of F [x]/(di) and

x

1x...

xei−1

= M(di)

1x...

xei−1

,where

M(di) =

0 10 1· ·· ·

0 1−ai,0 · · · · −ai,ei−1

is the companion matrix of di. 1, x, . . . , xei−1 correspond to an F -basis εi,1, . . . , εi,ei

of Vi. We have

T

εi,1...

εi,ei

= M(di)

εi,1...

εi,ei

.Now

⋃ri=1{εi,1, . . . , εi,ei} is an F -basis of V and

T

ε1,1...

ε1,e1...εr,1...

εr,er

=

M(d1)

. . .

M(dr)

ε1,1...

ε1,e1...εr,1...

εr,er

.

Hints for the Exercises

1.2. The following is the proof that g is onto. Proofs of other claims are eithersimilar or very simple.∀c ∈ C, γ(c) ∈ C ′,⇒ ∃b′ ∈ B′ such that g′(b′) = γ(c). Since g′′(β′(b′)) =

γ′(g′(b′)) = (γ′ · γ)(c) = 1, ∃a′′ ∈ A′′ such that f ′′(a′′) = β′((b′). Choosea′ ∈ A′ such that α′(a′) = a′′. Then β′(f ′(a′)) = f ′′(α′(a′)) = f ′′(a′′) =β′(b′), ⇒ b′[f ′(a′)]−1 = β(b) for some b ∈ B. Since γ(g(b)) = g′(β(b)) =g′(b′f ′(a′−1)) = g′(b′) = γ(c), g(b) = c.

1.6. Assume to the contrary that ∃H < An such that [An : H] = 2. Then ∀3-cycle σ ∈ An, σ2 ∈ H.

1.8. Label the vertices of the regular n-gon as 0, 1, . . . , n− 1. Then Dn = 〈α, β〉where α = (0, 1, . . . , n− 1) and β = (1, n− 1)(2, n− 2) · · · . For 0 ≤ i < n, αi

is a product of (i, n) disjoint n(i,n) -cycles; hence Fix(αi) = c(i,n). Note that

o(βαi) = 2. If n is odd, βαi fixes exactly one element of {0, . . . , n−1}; henceβαi is a product of n−1

2 disjoint transpositions.

1.9. Assume to the contrary that G is simple and |G| = 120 = 23 · 3 · 5. Let Pbe the set of all Sylow 5-subgroups of G. By Sylow’s theorem, |P| = 6. LetG act on P by conjugation. Since G is simple, the kernel of the action is Gor {e}. Rule out the possibility of G. Then G ↪→ SP = S6. So, may assumeG ⊂ S6. Since G is simple, G ⊂ A6. (Otherwise, [G : G ∩ A6] = 2 since|G|

|G∩A6| = |GA6||A6| ≤

|S6||A6| = 2.)

Let A6 act on A6/G by left multiplication. Since A6 is simple, A6 ↪→SA6/G = S3, →←.

For order 300, the proof is simpler. Only need the action of G on the setof Sylow 5-subsgroups of G

65

Solutions of the Exercises

1.4. Proof. 1◦ ker f = 1. ker f = ker(β · f) = ker(f ′ · α) = 1.2◦ g is onto. ∀c ∈ C, γ(c) ∈ C ′,⇒ ∃b′ ∈ B′ such that g′(b′) = γ(c). Since

g′′(β′(b′)) = γ′(g′(b′)) = (γ′ ·γ)(c) = 1, ∃a′′ ∈ A′′ such that f ′′(a′′) = β′((b′).Choose a′ ∈ A′ such that α′(a′) = a′′. Then β′(f ′(a′)) = f ′′(α′(a′)) =f ′′(a′′) = β′(b′), ⇒ b′[f ′(a′)]−1 = β(b) for some b ∈ B. Since γ(g(b)) =g′(β(b)) = g′(b′f ′(a′−1)) = g′(b′) = γ(c), g(b) = c.

3◦ im f ⊂ ker g. γ · g · f = g′ · f ′ · α = 1⇒ g · f = 1.4◦ ker g ⊂ im f . ∀b ∈ ker g, g′(β(b)) = γ(g(b)) = 1, ⇒ ∃a′ ∈ A′ such

that β(b) = f ′(a′). Since f ′′(α′(a′)) = β′(f ′(a′)) = β′ · β)(b) = 1, α(a′) = 1,↔ a ∈ A such that a′ = α(a). Now, β(f(a)) = f ′(α(a)) = f ′(a′) = β(b) ⇒b = f(a). �

1.5. Solution. Trivial subgroups: {id} and S4.Subgroup of order 12: A4 by Exercise 1.7.Subgroups of order 8: These are Sylow 2-subgroups. By Sylow’s theo-

rem, there are 3 of them. They are K〈(1, 2)〉, K〈(1, 3)〉, K〈(1, 4)〉, whereK = {id, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. These subgroups are also theconjugates of D4.

Subgroups of order 6: If H < S4 such that |H| = 6, then H contains anelement of order 3, which must be a 3-cycle. Then H must fix the fourthelement. (Otherwise, H would act transitively on {1, 2, 3, 4}, which wouldimply that 4

∣∣ |H|.) So, S must be the stabilizer of some i ∈ {1, 2, 3, 4}, i.e.,H = S3, (1, 4)S3(1, 4), (2, 4)S3(2, 4), (3, 4)S3(3, 4).

Subgroups of order 4: If cyclic, must be 〈(1, 2, 3, 4)〉, 〈(1, 2, 4, 3)〉, 〈(1, 3, 2, 4)〉.If not cyclic, must be K or generated by two disjoint transpositions, i.e.,〈(1, 2), (3, 4)〉, 〈(1, 3), (2, 4)〉, 〈(1, 4), (2, 3)〉.

Subgroups of order 3: 〈(1, 2, 3)〉, 〈(1, 2, 4)〉, 〈(1, 3, 4)〉, 〈(2, 3, 4)〉.Subgroups of order 2: 〈(i, j)〉, 1 ≤ i < j ≤ 4, and 〈(1, 2)(3, 4)〉, 〈(1, 3)(2, 4)〉,

〈(1, 4)(2, 3)〉.The normal subgroups are S4, A4, K, {id}. �

1.6. Proof. Assume to the contrary that ∃H < An such that [An : H] = 2.Then H C An. Then ∀ 3-cycle σ ∈ An, (σH)2 = H in An/H; hence σ2 ∈ H.Thus, σ = σ4 ∈ H. Since H contains all 3-cycles, H = An, →←. �

1.7. Proof. IfH 6= An, then [An : An∩H] = 2,→←. Can also use the argumentin the proof of Exercise 1.6. �

1.8. Solution. Label the vertices of the regular n-gon as 0, 1, . . . , n − 1. ThenDn = 〈α, β〉 where α = (0, 1, . . . , n− 1) and β = (1, n− 1)(2, n− 2) · · · . For0 ≤ i < n, αi is a product of (i, n) disjoint n

(i,n) -cycles; hence Fix(αi) = c(i,n).

67

68 SOLUTIONS OF THE EXERCISES

Note that o(βαi) = 2 and that (βαi)(k) = k ⇔ 2k ≡ −i (mod n). If n isodd, βαi fixes exactly one element of {0, . . . , n− 1}; hence βαi is a productof n−1

2 disjoint transpositions. If n is even and i is even, βαi fixes exactlytwo elements of {0, . . . , n − 1}; hence βαi is a product of n

2 − 1 disjointtranspositions. If n is even and i is odd, βαi fixes exactly no element of{0, . . . , n−1}; hence βαi is a product of n2 disjoint transpositions. Therefore,

Fix(βαi) =

c

n+12 if n is odd,

cn2 +1 if n is even and i even,c

n2 if n is even and i odd.

By Burnside’s lemma, the number of Dn-orbits in X is

12n

[n−1∑i=0

Fix(αi) +n−1∑i=0

Fix(βαi)]

=

{12n

∑d|n φ(nd )cd + 1

2cn+1

2 if n is odd,12n

∑d|n φ(nd )cd + 1

4 (c+ 1)cn2 if n is even.

�

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