MAS3219/MAS8219Groups,Graphs andSymmetry Assignment ...najd2/teaching/mas3219/assign.pdf · MAS3219/MAS8219Groups,Graphs andSymmetry Assignment Exercises: Solutions 1 Exercises forMonday

MAS3219/MAS8219 Groups, Graphs and Symmetry

Assignment Exercises: Solutions

1 Exercises for Monday 4th February

This assignment will be assessed at the problem class on 4th February.Please prepare solutions to the questions and be ready to present part of oneof the solutions to the class, on the board, during the problem class. Markswill be awarded as follows.For attendance: 1, for speaking at the board: 1, for making sense: 1, fora star performance: 1.Feedback on written solutions that you hand in at the problem class will beprovided.

1.1 Fill in the details of the 2 statements following Definition 1.23 in the notes. Includethe following.

(a) Suppose that I is a (possibly infinite) set and that, for all i ∈ I, we have asubgroup Hi of a group G. Then

H = ∩i∈IHi

is a subgroup of G. Using this fact explain (quickly) why 〈S 〉 in Definition 1.23is a group. From now on the question assumes the set up of Definition 1.23.

(b) Explain (also quickly) why S ⊆ 〈S 〉.(c) Explain (quickly again) why a subgroup containing S must also contain 〈S 〉.(d) Use the above to show that 〈S 〉 is the smallest subgroup containing S (including

a brief explanation of what “smallest” means in this context.)

(e) For s1, . . . , sn ∈ S and ε1, . . . , εn ∈ {±1} find the inverse of sε11 · · · sεnn (justifyingany claims you make).

(f) LetT = {sε11 · · · sεnn : n ≥ 0, si ∈ S, εi = ±1}.

Show, using Lemma 1.4, that T ≤ G. (If S is empty what does T contain?)

(g) Show that T = 〈S 〉.

Solution

(a) Use Lemma 1.4. As the identity element 1G of G is contained in all subgroupsof G, we have 1G ∈ Hi, for all i ∈ I, so 1G ∈ H . Therefore H is non-empty.

1

MAS3219/8219 Assignment Exercises: Solutions 2

Suppose x, y ∈ H . Then, for all i ∈ I, we have x, y ∈ Hi, so from Lemma 1.4we have xy−1 ∈ Hi. Therefore xy−1 ∈ ∩i∈IHi = H . From Lemma 1.4 again, His a subgroup of G.

As 〈S 〉 is an intersection of subgroups of G it now follows that 〈S 〉 is also asubgroup of G.

(b) By definition

∩{H ≤ G : H ⊇ S} = {x ∈ G : x ∈ H, for all H ≤ G such that H ⊇ S}

and we have〈S 〉 = ∩{H ≤ G : H ⊇ S}.

If s ∈ S then s ∈ H , for all subgroups of G containing S, so s ∈ ∩{H ≤ G |H ⊇S} = 〈S 〉. Hence S ⊆ 〈S 〉.

(c) Suppose H ≤ G and S ⊆ H . If x ∈ 〈S 〉 then, by definition, x ∈ H . Hence〈S 〉 ⊆ H .

(d) The “smallest subgroup containing S” is something like a least upper bound ofsubgroups containing S. More precisely, suppose M is a subgroup satisfying

(i) M contains S, and

(ii) if H is a subgroup such that S ⊆ H then M ⊆ H ,

then we say M is a minimal subgroup containing S. If M1 and M2 are minimalsubgroups containing S then S ⊆ M1 ∩M2, by condition (i). Therefore, by (ii),we have M1 ⊆ M1 ∩M2; so M1 ⊆ M2. Similarly, M2 ⊆ M1 and thus M1 = M2.This means that a minimal subgroup containing S is unique if it exists at all.As we have seen in (b) and (c) above, for any subset S the subgroup 〈S 〉satisfies both conditions (i) and (ii), so is a minimal subgroup containing S.The conclusion is that a minimal subgroup containing S exists, is unique, andis equal to 〈S 〉 (and this holds for any subset S of G). We can now say “thesmallest subgroup containing S”, instead of “a minimal subgroup containingS”.

(e) Consider the element s−εnn · · · s−ε1

1 of G. We have

(sε11 · · · sεnn )(s−εnn · · · s−ε1

1 ) = (sε11 · · · sεn−1

n−1 )sεnn s−εn

n (s−εn−1

n−1 · · · s−ε11 )

= (sε11 · · · sεn−1

n−1 )(s−εn−1

n−1 · · · s−ε11 )

...

= sε11 s−ε11

= 1G.

(Strictly speaking this is an induction argument.) To see that

(s−εnn · · · s−ε1

1 )(sε11 · · · sεnn ) = 1

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we can replace each εi by −εi and relabel. Thus

(sε11 · · · sεnn )−1 = s−εnn · · · s−ε1

1 .

(f) When n = 0 the sequence of length 0 is, by convention, the identity element ofG. Hence (even if S = ∅) the element 1G belongs to T ; and so T is non-empty.Let x = sε11 · · · sεmm and y = tδ11 · · · tδnn be elements of T , with m,n ≥ 0, si, ti ∈ S,and εi, δi ∈ {±1}. Then, from the above

y−1 = t−δnn · · · t−δ1

1 .

Thusxy−1 = sε11 · · · sεmm t−δn

n · · · t−δ11 ,

which is again an element of T . That is, xy−1 ∈ T . Hence T is a subgroup ofG.

(g) By definition every element of s belongs to T ; so S ⊆ T . Moreover from theprevious part of the question, T is a subgroup of G. Hence 〈S 〉 ⊆ T . Everyelement of the form sε11 · · · sεmm , with si ∈ S and εi = ±1, belongs to 〈S 〉, as〈S 〉 is a group, so T ⊆ 〈S 〉. Hence T = 〈S 〉.

1.2 (a) Give the definitions of the order of a group and the order of an element of agroup.

(b) A group is cyclic if it can be generated by a single element. Show that if gbelongs to the group G and has order n then {g} generates a cyclic subgroupof G of order n.

Solution

(a) The order of an element g of a group G is the least positive integer such thatgn = 1G.

(b) The group 〈 g 〉 generated by {g}, is cyclic by definition. From the previousquestion it consists of elements of the form tε1 · · · tεr , for some r ≥ 0 and εi = ±1.We may simplify tε1 · · · tεr to tm, where m =

∑r

i=1 εi, so 〈 g 〉 = {gm : m ∈ Z}.Let |g| = n, so n ≥ 1. For all m ∈ Z, we may write m = nq + r, where0 ≤ r < n. Hence we have gm = gnq+r = (gn)qgr = gr. Thus any element of 〈 g 〉may be written as gr, where 0 ≤ r < n. There are at most n such elements, so| 〈 g 〉 | ≤ n.

Finally we must show that no two elements gr and gs, with r 6= s, are equal.Assume that 0 ≤ r < s < n. Then gr = gs if and only if gs−r = 1G, whichimplies that |g| ≤ s − r. As s − r < n this is contrary to the assumption that|g| = n.

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2 Exercises for Friday 15th February

2.1 In Definition 2.1 of the notes the (external) direct sum K ⊕H of two groups (K, ◦)and (H,�) is defined. Verify the claim that the binary operation on K ⊕ H , givenin Definition 2.1, is associative.

(4 marks)

Solution. As (k, h)(k′, h′) = (k ◦ k′, h�h′) we have, for k, k′, k′′ ∈ K and h, h′, h′′ ∈H ,

((k, h)(k′, h′))(k′′, h′′) = (k ◦ k′, h�h′)(k′′, h′′)

= ((k ◦ k′) ◦ k′′, (h�h′)�h′′)

= (k ◦ (k′ ◦ k′′), h�(h′�h′′))

= (k, h)(k′ ◦ k′′, h′�h′′)

= (k, h)((k′, h′)(k′′, h′′)),

where the 3rd equality holds because ◦ and � are associative (that is because Axiom3 holds in K and H).

(4 marks)

2.2 This question considers a couple of generating sets for the direct sum of an infinitecyclic group and the symmetric group S3 of degree 3. First we establish a correspon-dence between two descriptions of the infinite cyclic group.

(a) Let C∞ denote the group with underlying set {xk : k ∈ Z} and binary operationxmxn = xm+n.

i. Show that C∞ is a group.

ii. Show that the map f : Z −→ C∞ given by f(m) = xm is a group isomor-phism.

(b) Let S3 be the symmetric group of degree 3 and write σ = (1 2 3) and τ = (1 2)as in the notes (page 6).

i. Show that the direct sum C∞ ⊕ S3 is generated by the elements (1, σ) and(0, τ).

ii. Now show that C∞ ⊕ S3 is generated by (3, σ) and (2, τ).

(13 marks)

Solution.

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(a) i. The operation given is a binary operation. The axioms for a group must bechecked.

Axiom 1. We have x0 ∈ C∞ and, for all xm ∈ C∞,

x0xm = xm = xmx0,

so x0 = 1C∞.

Axiom 2. If xm ∈ C∞ then x−m ∈ C∞ and

xmx−m = x0 = x−mxm,

so x−m = (xm)−1. Hence Axiom 2 holds.

Axiom 3. If xa, xb, xc ∈ C∞ then

(xaxb)xc = xa+bxc = xa+b+c = xaxb+c = xa(xbxc).

Hence Axiom 3 holds.

(3 marks)

ii. If m,n ∈ Z then

f(m+ n) = xm+n = xmxn = f(m)f(n),

so f is a homomorphism.If xm ∈ C∞ then xm = f(m), so f is surjective.If f(m) = 1 then xm = x0, so m = 0. Hence f is injective.Combining these facts, f is an isomorphism.

(2 marks)

(b) Here we identify C∞ with Z and S3 with the set {e, σ, σ2, τ, στ, σ2τ}.i. Let A be the subgroup of C∞ ⊕ S3 generated by the elements (1, σ) and

(0, τ). We must show that (m, ρ) ∈ A, for all m ∈ Z and ρ ∈ S3. Wehave immediately (1, σ), (0, τ), (0, e) = (1, σ)0 and (−1, σ2) = (1, σ)−1 inA. Then

(0, τ)(1, σ) = (1, τσ) = (1, σ2τ) ∈ A,

so(1, σ2τ)(0, τ) = (1, σ2) ∈ A.

Therefore(−1, σ2)(1, σ2) = (0, σ) ∈ A.

Now we have the elements

(0, σ2) = (0, σ)2, (0, στ) = (0, σ)(0, τ) and (0, σ2τ) = (0, σ2)(0, τ)

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in A, so A contains (0, ρ), for all ρ ∈ S3.For all m ∈ Z we have (1, σ)m = (m, σm) ∈ A, and (0, σ−m) = (0, σr) ∈ A,where 0 ≤ r < 3 and −m ≡ r (mod 3). Thus

(m, e) = (m, σm)(0, σ−m) ∈ A.

Thus (m, e) ∈ A, for all m ∈ Z. Hence (m, ρ) = (m, e)(0, ρ) ∈ A, for allm ∈ Z, ρ ∈ S3.

(5 marks)

ii. Now let B be the subgroup of C∞⊕S3 generated by (3, σ) and (2, τ). ThenB contains

(3, σ)3 = (9, e) and (2, τ)4 = (8, e)

so also(9, e)(8, e)−1 = (9, e)(−8, e) = (1, e).

Hence B contains (1, e)m = (m, e), for all m ∈ Z. Thus B contains

(−3, e)(3, σ) = (0, σ) and (−2, e)(2, τ) = (0, τ).

As in the previous case it follows that B contains (0, ρ) for all ρ ∈ S3, soB = C∞ ⊕ S3.

(3 marks)

2.3 Let Q denote the group of rational numbers under the binary operation of addition.This is an Abelian group.

(a) Show that every element of Q has infinite order.

(b) Show that Q is not finitely generated.

(c) Fix a prime p. Using Lemma 1.4, show that Qp = {m/pn : m,n ∈ Z, n > 0} isa subgroup of Q.

(d) Show that Z is a subgroup of Qp. (We know Z is a subgroup of Q, so it’s enoughto check that it is a subset of Qp.)

(11 marks)

Solution.

(a) Of course the question should say “Show that every non-trivial element of Qhas infinite order.” A non-trivial element of Q may be written in the form r/s,where r, s ∈ Z, s > 0, r 6= 0 and gcd(r, s) = 1. Suppose then that r/s is anon-trivial element of Q, written in this form, so r 6= 0. The order of r/s is theleast positive integer n such that n(r/s) = 0, if such an integer exists, and isinfinite otherwise. If n is a positive integer then n(r/s) = nr/s 6= 0, as r 6= 0.Hence r/s has infinite order.

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(2 marks)

(b) Let r1/s1, . . . , rn/sn be elements of Q, with si > 0, and let H = 〈r1/s1, . . . rn/sn〉be the subgroup they generate. We may assume that ri/si 6= 0, as the removalof 0 from the generating set of a subgroup leaves the subgroup unchanged. Lets = lcm(s1 · · · sn) (any positive common multiple of s1, . . . , sn will do) and letsi = s/si, for i = 1, . . . , n. Then s = sisi, so

ri/si = risi/sisi = risi/s,

for all i.

Now, an arbitrary element of H has the form

h = m1(r1/s1) + · · ·+mn(rn/sn),

for some mi ∈ Z (since Q is an Abelian group), so has the form

h = m1(r1s1/s) + · · ·+mn(rnsn/s) = (m1r1s1 + · · ·+mnrnsn)/s.

Thus every element of h can be written as a/s, for some integer a.

Now consider 1/(s+ 1) ∈ Q. If 1/(s+ 1) ∈ H there must be an integer a suchthat 1/(s + 1) = a/s, so s = a(s + 1). This implies s + 1 divides s, which inturn implies s+ 1 ≤ s, since s and s+ 1 are positive. This contradiction showsthat 1/(s + 1) /∈ H . We conclude that if H is a finitely generated subgroup ofQ then there exists an element of Q not contained in H . This means that Q isnot finitely generated.

(6 marks)

(c) The element 0/p ∈ Qp, so Qp 6= ∅. (Any element would do.) Let a/pk andb/pl be elements of Qp, where a, b, k, l ∈ Z, k, l > 0. Then −(b/pl) = −b/pl

and a/pk + (−b/pl) = (apl − bpk)/pk+l ∈ Qp. Hence, from Lemma 1.4, Qp is asubgroup of Q.

(2 marks)

(d) For all m ∈ Z we have m = mp/p ∈ Qp, so Z ⊆ Qp. As Z is a subgroup ofQ it follows that Z is a subgroup of Qp. (Both statements of Lemma 1.4 hold,because Z ≤ Q.)

(1 mark)

2.4 Let C∗ denote the group of non-zero complex numbers C\{0} under multiplication.This is an infinite Abelian group. Fix a prime p and let Z(p∞) = {exp(2πim/pn) ∈C∗ : m,n ∈ Z, n > 0}.

(a) Show that Z(p∞) is a subgroup of C∗. (This group is called the Prufer group orp-quasicyclic group.)

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(b) Show that every element of Z(p∞) has (finite) order a power of p.

(c) Show that every finitely generated subgroup of Z(p∞) is a finite cyclic group.

(d) Show that Z(p∞) is not finitely generated.

(e) Show that Z(p∞) ∼= Qp/Z (the quotient of the group Qp of the previous questionby the integers Z).

(22 marks)

Solution.

(a) Setting m = n = 0 we have exp(0) = 1 ∈ Z(p∞), so Z(p∞) 6= ∅.Suppose that x = exp(2πia/pk) and y = exp(2πib/pl) are elements of Z(p∞),where a, b, k, l ∈ Z, k, l > 0. Then y−1 = (exp(2πib/pl))−1 = exp(−2πib/pl) and

xy−1 = exp(2πia/pk) exp(−2πib/pl)

= exp([2πia/pk]− [2πib/pl])

= exp(2πi(apl − bpk)/pk+l) ∈ Z(p∞).

Hence, from Lemma 1.4, Z(p∞) is a subgroup of C∗.

(3 marks)

(b) Let x = exp(2πia/pk) be an element of Z(p∞). Then

xpk = (exp(2πia/pk))pk

= exp(2πiapk/pk) = exp(2πia) = 1 ∈ C∗.

Hence the order of x is at at most pk and if x has order d then d|pk. As p is primethis means d is a power of p, as required. [The following is a standard result ingroup theory. If an element g of a group satisfies gn = 1 then the order of the

element g divides n. To see this, suppose the order of g is d > 0. Then we maywrite n = dq+r, where 0 ≤ r < d. Thus 1 = gn = gdq+r = gdqgr = (gd)qgr = gr,as gd = 1. Hence gr = 1, and as d is the smallest positive integer such thatgd = 1, it follows that r = 0. Hence n = qd, as claimed.]

(2 marks)

(c) The solution is by induction, but it pays to begin by considering the first twocases. By definition a single element of Z(p∞) generates a cyclic group, and thisis the base case for the induction. Now consider the subgroup generated by twonon-trivial elements x = exp(2πia/pk) and y = exp(2πib/pl) of Z(p∞). Withoutloss of generality we may assume k ≤ l (otherwise swap x and y). Moreover, wemay assume that p ∤ a and p ∤ b. Every element t of 〈x, y〉 has the form xmyn,

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for some m,n ∈ Z, as Z(p∞) is an Abelian group. Therefore, if t ∈ 〈x, y〉 then,for some m,n ∈ Z,

t = (exp(2πia/pk))m(exp(2πib/pl))n

= exp(2πiam/pk) exp(2πibn/pl)

= exp(2πi(ampl + bnpk)/pk+l)

= exp(2πipk(ampl−k + bn)/pk+l)

= exp(2πi(ampl−k + bn)/pl),

as k ≤ l. Now let S = {s = m(apl−k) + nb ∈ Z : m,n ∈ Z}. From stage 1number theory, S = {qd ∈ Z : d = gcd(apl−k, b), q ∈ Z}. Hence we have shownthat if t ∈ 〈x, y〉 then t = exp(2πis/pl), where s = qd, for some q ∈ Z, andd = gcd(apl−k, b). Conversely, if t = exp(2πis/pl), with s ∈ S, then s = qd =mapl−k + nb, for some m,n ∈ Z. Reversing the argument above, we then havet = xmyn, so t ∈ 〈x, y〉. Hence

〈x, y〉 = {exp(2πidq/pl) : q ∈ Z} = 〈z〉,

wherez = exp(2πid/pl).

Thus, any subgroup of Z(p∞) generated by two elements is cyclic.

Now for the induction. Assume that, for some k ≥ 1, any subgroup generatedby k elements is cyclic. Let H be generated by elements x1, . . . , xk+1 of Z(p∞).Then, from the inductive assumption, H ′ = 〈x1, . . . , xk〉 is a cyclic group, sayH ′ = 〈u〉, for some u ∈ Z(p∞). Moreover H = 〈H ′, xk+1〉 = 〈u, xk+1〉, and soby the argument above, for two elements, H is also cyclic. As all elements ofZ(p∞) have finite order, H is a finite cyclic group.

(7 marks)

(d) Let H be a finitely generated subgroup of Z(p∞). From (c), H = 〈z〉, for somez = exp(2πid/pn) ∈ Z(p∞). Consider the element v = exp(2πi/pn+1) of Z(p∞).If v ∈ H then v = zm, for some m ∈ Z, so

exp(2πi/pn+1) = exp(2πidm/pn),

which implies that

exp(2πi(1− dmp)/pn+1) = 1 ∈ C.

Therefore, it must be that pn+1|(1− dmp), so 1− dmp = pn+1q, for some q ∈ Z,and we obtain 1 = p(pnq+dm), so p|1, a contradiction. This shows that v /∈ H ,and so no finitely generated subgroup contains all elements of Z(p∞). HenceZ(p∞) is not finitely generated.

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(4 marks)

(e) We shall show that there is a surjective homomorphism from Qp onto Z(p∞)with kernel Z. Define θ : Qp −→ Z(p∞) by

θ(m/pn) = exp(2πim/pn).

To see that this is a homomorphism let a/pk and b/pl be elements of Qp. Then

θ((a/pk) + (b/pl)) = θ((apl + bpk)/pk+l)

= exp(2πi(apl + bpk)/pk+l)

= exp(2πiapl/pk+l) exp(2πibpk/pk+l)

= exp(2πia/pk) exp(2πib/pl)

= θ(a/pk)θ(b/pl).

Hence θ is a homomorphism.

If x = exp(2πim/pn) ∈ Z(p∞) then x = θ(m/pn), so θ is surjective.

Hence Z(p∞) ∼= Qp/ ker(θ), and it remains to find ker(θ). We have m/pn ∈ker(θ) if and only if exp(2πim/pn) = 1, if and only if pn|m if and only ifm = pnq, for some q ∈ Z, if and only if m/pn ∈ Z. Therefore ker(θ) = Z.Thus

Z(p∞) ∼= Qp/Z,

as claimed.

(6 marks)

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3 Exercises for Friday 1st March

3.1 Let G be a group with subgroups H and K such that G is generated by the setH ∪K. Show that if k−1hk ∈ H , for all h ∈ H and k ∈ K, then H ⊳ G.

(4 marks)

Solution Let h1, h2 ∈ H . Then h−11 h2h1 ∈ H , so for all h ∈ H and x ∈ H ∪K we

have x−1hx ∈ H .

If g is an arbitrary element of G then g = xε11 · · ·xεn

n , for some xi ∈ H ∪ K andεi = ±1. As xi ∈ H ∪K if and only if x−1

i ∈ H ∪K we can assume εi = 1, for alli. Therefore, for all g ∈ G we have g = x1 · · ·xn, with xi ∈ H ∪K, for some n ≥ 0.Then

g−1hg = x−1n · · ·x−1

1 hx1 · · ·xn = x−1n · · ·x−1

2 h1x2 · · ·xn = · · · = x−1n hn−1xn = hn ∈ H,

where h1 = x−11 hx1 and hi = x−1

i hi−1xi ∈ H , for i = 2, . . . n. Hence H ⊳ G.

3.2 Let G be a group with subgroups N and K, such that N is normal in G.

(a) Show, using Lemma 1.4, that

NK = {nk : n ∈ N, k ∈ K}

is a subgroup of G.

(2 marks)

(b) Let H denote the subgroup of G generated by N ∪K. Show, using the remarksfollowing Definition 1.23 in the notes, that NK ≤ H .

Use this to show that H = NK.

(3 marks)

Solution

(a) As NK 6= ∅ we must only check that if n1k1 and n2k2 belong to NK, withni ∈ N and ki ∈ K, then (n1k1)(n2k2)

−1 ∈ NK. We have

(n1k1)(n2k2)−1 = n1k1k

−12 n−1

2 = n1(k1k−12 )n−1

2 (k1k−12 )−1(k1k

−12 ) = n3k3,

where n3 = (k1k−12 )n−1

2 (k1k−12 )−1 ∈ N , as N ⊳ G, and k3 = k1k

−12 ∈ K. Thus

(n1k1)(n2k2)−1 ∈ NK, which is therefore a subgroup of G.

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(b) From the remarks in the notes, the subgroup H generated by N ∪K consists ofelements of the form sε11 · · · sεnn , where si ∈ N ∪K and εi = ±1. Every elementof NK is of this form, so NK ≤ H . However N ⊆ NK and K ⊆ NK, aselements n of N can be written as n · 1, with 1 ∈ K and elements k of K as1 · k. Hence N ∪K ⊆ NK. As H is the smallest subgroup containing N ∪Kwe must then have H ≤ NK. Hence H = NK.

3.3 Let Γ be the graph shown below.

∗

0 1

00 01 10 11

This question will determine the structure of Sym(Γ), the group of isomorphisms ofΓ. By considering the degrees of vertices we can see that every isomorphism of Γmaps ∗ to ∗, maps {0, 1} to {0, 1} and permutes the leaves of the graph. First it’snecessary to set up some notation. For a start, whenever describing the effect of anisomorphism f on vertices we assume that f(∗) = ∗, without further mention.

Let Id denote the identity map of Γ and let σ denote the map such that

σ(0) = 1, σ(1) = 0, σ(0j) = 1j, σ(0j) = 1j.

Denote by τL the isomorphism of Γ such that

τL(0) = 0, τL(1) = 1, τL(00) = 01, τL(01) = 00, τL(1j) = 1j.

That is, τL fixes all vertices except those below 0, which are swapped. Similarly, letτR be the isomorphism of Γ such that

τR(0) = 0, τR(1) = 1, τR(0j) = 0j, τR(10) = 11, τR(11) = 10.

Then σ, τL and τR are isomorphisms of Γ. As (τL)2 = (τR)2 = Id it is also true thatboth

L = {Id, τL} and R = {Id, τR}are subgroups of Sym(Γ), both of which are cyclic of order 2.

(Also, isomorphisms of this graph are completely determined by their effect on ver-tices, so to check that two of them are the same it’s sufficient to check their effectson vertices.)

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(a) Show that if ρ1 ∈ L and ρ2 ∈ R then

ρ1ρ2 = ρ2ρ1.

[Consider the effect of the lhs and rhs on vertices.]

Use this to show that LR = {ρ1ρ2 : ρ1 ∈ L, ρ2 ∈ R} is a subgroup of Sym(G).[This is similar to, but not the same as, Question 3.2(a).]

(4 marks)

(b) By checking the necessary conditions, show that LR is the internal direct sumof L and R and then, by quoting an appropriate theorem, that LR ∼= L⊕ R.

(2 marks)

(c) Show that the subgroup T = 〈σ〉 of Sym(Γ) is cyclic of order 2.

(2 marks)

(d) Show that LR is normal in the subgroup S of Sym(Γ) generated by the union ofLR and T . [It’s sufficient to show that if x ∈ T and y ∈ LR then x−1yx ∈ LR.Consider the effect of x−1yx on the various different types of vertex.]

(8 marks)

(e) Show that (LR) ∩ T = {Id}.(1 mark)

(f) By checking that an appropriate theorem from the notes applies, show thatS ∼= (LR)⋊ T .

(2 marks)

(g) Explain why Sym(Γ) = S and hence that Sym(Γ) ∼= (L ⊕ R) ⋊ T . [To seethat every element of Sym(Γ) belongs to S consider the effect of an element onvertices. Break the argument into two cases. First suppose your element fixes0. Then consider what happens if it swaps 0 and 1: try composing it with σ.]

(7 marks)

Solution

(a) If ρ1 = Id then ρ1ρ2 = ρ2 = ρ2ρ1, as required. Similarly, if ρ2 = Id then theresult holds.

If ρ1 = τL and ρ2 = τR then

ρ1ρ2(j) = ρ1(j) = j = ρ2(j) = ρ2ρ1(j),

AJD January 2013


for j ∈ {0, 1}, and, for x, y ∈ {0, 1},

ρ1ρ2(xy) = ρ1(x[1 − y]) = [1− x][1 − y],

whileρ1ρ2(xy) = ρ1([1− x]y) = [1− x][1 − y].

Hence ρ1ρ2 = ρ1ρ2, for all ρ1 ∈ L and ρ2 ∈ R.

If f1g1 and f2g2 are elements of LR, where fi ∈ L and gi ∈ R, then

(f1g1)(f2g2)−1 = f1g1g

−12 f−1

2 = f1f−12 g1g

−12 ∈ LR,

where the second equality holds because elements of L commute with elementsof R, as we have just shown. From Lemma 1.4, as LR is not empty it is asubgroup of Sym(G).

(4 marks)

(b) We must check the conditions of Definition 2.3. We have just verified condition3. Condition 1 holds by definition of LR. All that remains to be done is tocheck that L ∩ R = {Id}. If f ∈ L then f(1j) = 1j. As τR(10) = 11 it followsthat if f is also in R then f = Id. Hence L∩R = {Id}, and Theorem 2.6 impliesthat LR ∼= L⊕R.

(2 marks)

(c) We must show that σ2 = Id. As σ fixes all vertices except those of the form ij,i, j ∈ {0, 1}, we need only check the effect of σ2 on such vertices. We have

σ2(0j) = σ(1j) = 0j and σ2(1j) = σ(0j) = 1j.

Hence σ2 = Id and 〈σ〉 = {Id, σ}, a cyclic group of order 2.

(2 marks)

(d) Let x ∈ T and y ∈ LR. If x = Id then certainly x−1yx = y ∈ LR. Hence wemay assume x = σ. As LR is the internal direct sum of L and R its elementsare Id, τL, τR and τLτR. We have σ−1 = σ, as σ2 = Id, so

σ−1τLσ(0) = σ−1τL(1) = σ−1(1) = 0 and σ−1τLσ(1) = σ−1τL(0) = σ−1(0) = 1.

Similarly σ−1τRσ(0) = 0 and σ−1τRσ(1) = 1.

For y ∈ {0, 1},

σ−1τLσ(0y) = σ−1τL(1y) = σ−1(1y) = 0y,

andσ−1τLσ(1y) = σ−1τL(0y) = σ−1(0[1− y]) = 1[1− y],

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henceσ−1τLσ = τR.

We deduce thatσ−1τRσ = σ−1(σ−1τLσ)σ = τL

and then that

σ−1τLτRσ = σ−1τLσσ−1τRσ = τRτL = τLτR.

Therefore, for all y ∈ LR, we have σ−1yσ ∈ LR, and so LR is normal in thesubgroup S.

(8 marks)

(e) As σ(0) = 1 and f(0) = 0, for all f ∈ LR, the isomorphism σ /∈ LR. HenceLR ∩ T = {Id}.

(1 mark)

(f) We have shown that LR ⊳ S and that LR ∩ T = {Id}. To apply Theorem 4.11we need to check that S = (LR)T . This follows from the result of Question3.2(b). Theorem 4.11 now implies that S ∼= (LR)⋊ T .

(2 marks)

(g) Every element of S is an isomorphism of Γ. On the other hand if f is anisomorphism of Γ then, as noted in the preamble, f fixes ∗, maps {0, 1} to{0, 1} and permutes the leaves of Γ. If f(0) = 0 then f(1) = 1 and f doesnothing except permute the leaves. As 00 and 01 are incident to 0 and not to 1,both f(00) and f(01) must be incident to f(0) = 0. Similarly, f(10) and f(11)must be incident to 1. Thus f maps {00, 01} to itself and {10, 11} to itself.Since f is a bijection, it can only be an element of LR.

Now, if f(0) = 1 and f(1) = 0 we can compose f with σ to give f ′ = σ◦f . Thenf ′ ∈ Sym(G) and f ′(0) = 0. Hence f ′ ∈ LR, using the result of the previousparagraph. Therefore f ∈ T (LR) = (LR)T = S. Thus, in all cases f ∈ S, soSym(G) = S.

(7 marks)

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4 Exercises for Friday 15th March

4.1 Show that the isometry of R2 given by

f(x, y) =

(

3−√3− x

2−

√3y

2,−3−

√3 +

√3x

2− y

2

)

is a direct isometry and a rotation and express it in the form (v, A) for appropriatev and A. Find its centre and angle of rotation.

(10 marks)

Solution

f(0, 0) =(

3−√3,−3−

√3)

,

so

v =

(

3−√3

−3−√3

)

.

f(x, y)− f(0, 0) =

(

−x

2−

√3y

2,

√3x

2− y

2

)

,

so

Ax =

(

−x2−

√3y2

√3x2

− y

2

)

=

(

−12

−√32

√32

−12

)(

x

y

)

and

A =

(

−12

−√32

√32

−12

)

.

(3 marks)

Thus f = (v, A). As det(A) = 14+ 3

4= 1, f is direct, and as A 6= I2, f is a rotation.

(2 marks)

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c = (I2 − A)−1v =

(

32

√32

−√3

232

)−1

v

=1

94+ 3

4

(

32

−√3

2√32

32

)

v

=

(

12

−12√3

12√3

12

)

v =1

2

(

1 − 1√3

1√3

1

)

v

=1

2

(

3−√3 +

√3 + 1

√3− 1− 3−

√3

)

=

(

2

−2

)

.

Thus c = (2,−2)t.

(4 marks)

[Check: f(2,−2) = (2,−2), so this really is the centre of rotation.]

A =

(

cos(θ) − sin(θ)sin(θ) cos(θ)

)

,

where θ = 2π3, so f is rotation through 2π

3radians, anticlockwise about (2,−2).

(1 mark)

4.2 Show that the isometry (v, B) of R2, given by

g(x, y) =

(

4x

5− 3y

5+ 2,−3x

5− 4y

5+ 1

)

is an opposite isometry and a glide reflection and express it in the form (2a+b, B) forappropriate a,b and B. Find the equation of its axis of reflection and the distanceof its minimal translation parallel to this axis.

(12 marks)

Solutiong(0, 0) = (2, 1)

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so

v =

(

21

)

and

g(x, y)− g(0, 0) =

(

4x

5− 3y

5,−3x

5− 4y

5

)

so

Bx =

(

4x5− 3y

5

−3x5− 4y

5

)

=

(

45

−35

−35

−45

)(

x

y

)

and

B =

(

45

−35

−35

−45

)

=1

5

(

4 −3

−3 −4

)

.

(3 marks)

Thus g = (v, B) and det(B) = 125(−16− 9) = −1, so g is opposite.

Bv =1

5

(

4 −3

−3 −4

)(

2

1

)

=1

5

(

5

−10

)

=

(

1

−2

)

6= −v,

so g is a glide reflection.

(2 marks)

a =1

4(v − Bv) =

1

4

((

2

1

)

−(

1

−2

))

=

(

14

34

)

and

b = v − 2a =

(

2

1

)

−(

12

32

)

=

(

32

−12

)

.

(3 marks)

[Check: B should satisfy Ba = −a and Bb = b:

Ba =1

5

(

4 −3

−3 −4

)(

14

34

)

=1

5

(

−54

−154

)

=

(

−14

−34

)

= −a,

and

Bb =1

5

(

4 −3

−3 −4

)(

32

−12

)

=1

5

(

152

−52

)

= b.

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As v = 2a+ b, we have the required expression for g.]

The axis of reflection of g is a line in direction of the vector b and passing throughthe point a. The line through the origin in direction of b is the set of points

sb = s

(

32

−12

)

, for s ∈ R,

that is

s

(

3

−1

)

, s ∈ R.

Therefore the axis of reflection consists of points

sb+ a = s

(

3

−1

)

+

(

14

34

)

=

(

3s+ 14

−s + 34

)

, s ∈ R.

These are points (x, y)t with

x = 3s+1

4

y = −s+3

4

which is equivalent to

x− 1

4= 3s

y − 3

4= −s

or

−1

3

(

x− 1

4

)

= y − 3

4.

Rearranging this we find that the axis of reflection is the line m with equation

y = −x

3+

5

6.

[Alternatively apply the standard formula: the line parallel to b and passing througha has equation b0(y−a1) = b1(x−a0), that is

32(y− 3

4) = −1

2(x− 1

4). After rearranging

this gives the same answer.]

The vector b has length ||b|| =√

94+ 1

4=√

104, so b is a translation parallel to m

through a distance of√

52.

AJD January 2013


(4 marks)

4.3 This question completes the detail of one case of the classification of wallpaper groups:namely ∗2222.Let W be a wallpaper group and let L and O be its lattice and point group, re-spectively. Assume that L is rectangular so L = {αa + βb |α, β ∈ Z}, for fixedvectors a = (x, 0)t and b = (0, y)t, where we may assume 0 < x < y in R, andthat O = {I,−I, B0, Bπ}, where I is the identity matrix (so −I is the matrix ofrotation through π) and B0 and Bπ are the matrices of reflection in the x and y axes,respectively.

Assume further that W contains reflections g and h in lines parallel to the x and yaxes, respectively.

As g is reflection in a line parallel to the x-axis it is of the form (v, B0), for somevector v such that B0v = −v. Assume that v = αa+ βb, for some α, β ∈ R. Then

−v = B0v = αB0a+ βB0b = αa− βb

if and only ifαa− βb = −αa− βb

if and only if α = 0. Thus (v, B0) is a reflection if and only if v = βb, for someβ ∈ R. Writing b = 2β, we conclude that

g = (2bb, B0) for some b ∈ R.

(a) Use a similar argument to show that h = (2aa, Bπ), for some a ∈ R.

(2 marks)

Now, choose coordinate axes so that the x-axis is the axis of reflection of g and they-axis is the axis of reflection of h. Then W contains the reflections σ = g = (0, B0)and τ = h = (0, Bπ).

Consider the coset (L× {I})σ of L× {I} in W . We have

(L× {I})σ = {(αa+ βb, I2)(0, B0) : α, β ∈ Z}= {(αa+ βb, B0) : α, β ∈ Z}.

Thus(βb, B0) ∈ (L× {I})σ ⊆ W,

for all β ∈ Z, and (βb, B0) is a reflection (as B0βb = −βb) with axis parallel to thex-axis and passing through the point β

2b.

That is, W contains a reflection with axis parallel to the x-axis and passing throughthe point (β/2)b, for all β ∈ Z.

(L × {I})σ also contains elements (αa + βb, B0) with α 6= 0, and these are glidereflections with axis parallel to the x-axis, passing through the point β

2b, and with

translation through a distance of α in the direction of the x-axis.

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(b) Similarly, by considering (L × {I})τ , show that W contains a reflection withaxis parallel to the y-axis and passing through the point (α/2)a, for all α ∈ Z.

(2 marks)

(c) As above, give a geometric description of all symmetries in (L × {I})τ otherthan those described in the previous part of the question.

(2 marks)

(d) By considering the composition of σ and τ show that W contains the rotationρ = (0,−I).

(1 mark)

(e) By considering (L × {I})ρ, show that W contains a rotation through π withcentre of rotation (αa+ βb)/2, for all α, β ∈ Z.

(4 marks)

Using Lagrange’s theorem [W : (L × {I})] = |O|. Given this fact we may write Was a union of cosets of L× {I}: that is

W = (L× {I}) ∪ (L× {I})σ ∪ (L× {I})τ ∪ (L× {I})ρ.

Therefore we have now found all elements of W .

(f) Draw a diagram showing at least 9 points of L, vectors 0, a and b, axes ofreflection and centres of rotation, and marking one instance of each distincttype of centre of rotation and axis of reflection with an appropriate symbol.

(4 marks)

(g) Sketch a wallpaper pattern with symmetry group W .

(3 marks)

This is wallpaper group ∗2222.

(18 marks)

Solution

(a) h is reflection in a line parallel to the y-axis so it is of the form (u, Bπ), for somevector u such that Bπu = −u. If u = αa+ βb then

Bπu = αBπa+ βBπb = −αa+ βb = −u

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if and only if−αa+ βb = −αa− βb

if and only if β = 0. Writing a = 2α, we conclude that

h = (2aa, Bπ) for some a ∈ R.

(2 marks)

(b) Consider the coset (L× {I})τ of L× {I} in W . We have

(L× {I})τ = {(αa+ βb, Bπ) : α, β ∈ Z}.

Thus, for all β ∈ Z, W contains the reflection (αa, Bπ) with axis parallel to they-axis and passing through the point α

2a.

(2 marks)

(c) (L×{I})τ also contains elements (αa+βb, Bπ) with β 6= 0, and these are glidereflections with axis parallel to the y-axis, passing through the point α

2a, and

with translation through a distance of β in the direction of the y-axis.

(2 marks)

(d)στ = (0, B0)(0, Bπ) = (0, Aπ) = (0,−I) = ρ.

(1 mark)

(e) (L × {I})ρ = {(αa + βb,−I) : α, β ∈ Z}. The isometry (αa + βb,−I) is arotation through π about

c = (I − (−I))−1(αa+ βb)

=1

2I(αa+ βb)

=α

2a+

β

2b.

Thus W contains rotations through π about (αa+ βb)/2, for all α, β ∈ Z.

(4 marks)

(f)

AJD January 2013


a

b

2

2 2

2

• There are two orbits of horizontal axes of reflection: dotted-and-dashed-redand dashed-green.

• There are two orbits of vertical axes of reflection: dotted-and-dashed-yellowand dashed-black.

• There are four orbits of centres of rotation, all on axes of reflection. Onerepresentative of each orbit is marked with a solid blue disk and labelled 2.Other centres of rotation are marked with a blue circle.

(4 marks)

(g)

(3 marks)

AJD January 2013


5 Exercises for Friday 26th April

5.1 Draw Cayley graphs for the following groups G with generating sets S.

(a) G = D4, S = {σ, τ} as on pages 9–11 of the notes.

(2 marks)

(b) G = D∞, S = {σ, τ}. D∞ is the group with elements {σn, σnτ : n ∈ Z}, whereτ 2 = 1 and σ−1τ = τσ. Draw a diagram showing the vertices σn and σnτ , for−3 ≤ n ≤ 3, and all incident edges.

(2 marks)

(c) G = Z × Z, with binary operation addition, S = {x = (1, 1), y = (1,−1), z =(1, 2)}. Show vertices (m,n) for −1 ≤ m,n ≤ 2.

(2 marks)

5.2 Let G be a group acting on a graph Γ.

(a) Show that if G acts freely on Γ then G acts faithfully on Γ.

(2 marks)

(b) Exhibit an example to show the converse does not always hold. (You can useexamples from the notes, if any are suitable.)

(2 marks)

(c) Now let H be a subgroup of G. Show that if G acts faithfully on Γ then so doesH .

(2 marks)

(d) Show that if G acts freely on Γ then so does H .

(2 marks)

(e) Exhibit examples to show the converses of the two previous statements do notalways hold. (You can use examples from the notes, if any are suitable.)

(4 marks)

5.3 LetG be a group and S a generating set forG. Let Γ be the Cayley graph Γ(G, S) ofGwith respect to S. Let Sym+(Γ) be the symmetry group of the labelled, directed graphΓ. From the notes we know that G acts faithfully (in fact freely) on Γ. Thereforewe have an injective homomorphism α : G −→ Sym+(Γ). In this question we shallshow that α is surjective, so G ∼= Sym+(Γ). To do this suppose that f is an elementof Sym+(Γ). Then we shall show that f = α(g), for some g ∈ G. To keep thingsstraight, for all g ∈ G, we shall denote by vg the vertex g of Γ and by αg the image

AJD January 2013


α(g) of g in Sym+(Γ). This notation means that for all g, h ∈ G the action of g onthe vertex h is represented as αg(vh) = vgh, instead of gh, and that the edges of Γincident to vh are (vh, vhs), for s ∈ S.

Suppose then that f ∈ Sym+(Γ). Then f(v1G) = vg, for some g ∈ G.

(a) Show that αg−1 ◦ f(v1G) = v1G .

(1 mark)

(b) Next we shall show that any symmetry of Γ which fixes v1G is the identity mapon Γ. To do this suppose that φ ∈ Sym+(Γ) is such that φ(v1G) = v1G .

The edges incident to v1G are those of the form (v1G , vs) and (vs−1 , v1G), fors ∈ S (i.e. in the usual notation (1G, s) and (s−1, 1G)).

Show first that if (v1G , vs) is an edge incident to v1G then φ(vs) = vs. [This isan edge labelled s. How many of these are incident to v1G?]

Do the same for edges (vs−1 , v1G). Conclude that all vertices and edges incidentto v1G are fixed by φ.

Now argue by induction on the distance of vertices from v1G , that φ(vg) = vg,for all g ∈ G.

(8 marks)

(c) Using the previous two parts of the question show that f = αg = α(g). [Youwill need to show that αg−1 = α−1

g .] Complete the proof that G ∼= Sym+(Γ).

(3 marks)

AJD January 2013


6 Exercises for Friday 10th May

6.1 Apply the Stallings folding algorithm to find a free basis for each of the followingsubgroups of F (x, y, z). Show all steps of the algorithm and write out a free basisand the rank of the subgroup.

(a) H1 = 〈 yz, y−1z, z−1y 〉.(b) H2 = 〈xzx−2, xyzx−1z−1x−1, xyz−1y−1x−1 〉.(c) H3 = 〈xyz, xzyz, xyxz, xyzy 〉.

6.2 (a) Show that the subgroup of F (x, y) generated by x−1y−1xy and x−1y−2xy2 hasrank 2.

(b) Generalise this to show that, for all n ≥ 1, the free group F (x, y) contains afree group of rank n, generated by {x−1y−ixyi : 1 ≤ i ≤ n}. [Use an argumentbased on Stallings foldings.]

(c) Let Z = {zi : i ∈ Z, i ≥ 0} be a set (if i 6= j then zi 6= zj) and let f : Z −→F (x, y) be given by f(zi) = x−1y−ixyi, for all i ≥ 1.

Show, quoting and using an appropriate theorem, that there is a unique homo-morphism φ : F (Z) −→ F (x, y) with φ(zi) = f(zi), for all i ≥ 1.

(d) Show that ker(φ) = {1}. [Hint. If w is a reduced word over Z ∪Z−1 then thereis a positive integer n such that w is contained in the subgroup generated byz1, . . . , zn.]

Deduce that F (x, y) contains a free group of infinite rank.

6.3 Baumslag-Solitar groups. The groups with presentations

BS(m,n) = 〈a, b | a−1bma = bn〉,

for positive integers m,n are called Baumslag-Solitar groups (as they were first inves-tigated by Graham Higman). Here we consider BS(1, 2), the group with presentation

〈a, b | a−1ba = b2〉.

First we shall show that every element of BS(1, 2) can be written as an element ofthe set

L = {an : n ∈ Z} ∪ {a−kb2m+1ak+n : k,m, n ∈ Z}.This can be done by induction on the length of words in the generators. First it helpsto establish the following fact for elements of BS(1, 2).

(a) Show, by induction on r, that for all r ≥ 1,

a−rbar = b2r

in BS(1, 2).

AJD January 2013


Now for the proof that every element of BS(1, 2) can be represented by and elementof L. The unique word of length 0 is the empty word, and this represents the identityelement which can be written as a0, so is in L. Assume that all words in F (a, b), oflength at most n, are equal in BS(1, 2) to elements of L. To complete the inductivestep we must show that if w is a word of length n + 1 then it is equal, in BS(1, 2),to an element of L. Suppose then that w = xv, where x ∈ {a±1, b±1} and v isrepresented by an element of L. We check each possible form for v, for each possiblevalue of x. The possible forms of v are v = an or v = a−kb2m+1ak+n.

(b) Suppose that v = an. Check that a±1v and b±1v are elements of L. (No morethan three lines required.)

(c) Suppose that v = a−kb2m+1ak+n and x = a±1. Show (two lines only) that xv isin L.

(d) Suppose that v = a−kb2m+1ak+n, k ≥ 0 and x = b±1. (The argument whenk < 0 is given below.) Show that xv can be represented by an element of L, inthe case where k ≥ 0. [Hint. Replace a−kb2m+1ak+n by a power of b, using thefirst part of the question. It may help to remember that for all group elementsg, h it’s true that g−1hkg = (g−1hg)k.]

(In part 3d above the case k < 0 was omitted. It works a little differently to the casek ≥ 0. (Thanks to David Robertson for spotting this). If v = a−kb2m+1ak+n, wherek < 0, then, for ε = ±1,

bεv = bεa−kb2m+1ak+n

= (a−kak)bεa−kb2m+1ak+n

= a−k(akba−k)εb2m+1ak+n

= a−k(b2k

)εb2m+1ak+n( as −k > 0)

= a−kb2k×ε+2m+1ak+n

= a−kb2(2k−1×ε+m)+1)ak+n

= a−kb2M+1ak+n,

where M = 2k−1 × ε+m ∈ Z. Hence bεv has the required form in this case too. )

This completes the inductive step, so now we have shown that every element ofBS(1, 2) can be written as an element of L. Next we shall show that BS(1, 2) actson the real line. Consider the two maps α and β of R to itself given by

α(x) = x/2

andβ(x) = x+ 1.

Together α and β generate a subgroup A of the group of invertible maps from R toR, under composition of maps: that is A = 〈α, β〉. Define a map f : {a, b} −→ A byf(a) = α and f(b) = β.

AJD January 2013


(e) Use von Dyck’s Theorem to show that there is a homomorphism θ from BS(1, 2)to A such that θ(a) = α and θ(b) = β.

This means that we have an action of BS(1, 2) on R. Next we shall show that in factBS(1, 2) ∼= A.

(f) Is θ surjective and if so why?

(g) If g ∈ ker(θ) then, in particular θ(g) maps 0 to 0 and 1 to 1. Moreover g isrepresented by an element of L. Show that the image of 1 under θ(an) is not 1,unless n = 0. Show that the image of 0 under θ(a−kb2m+1ak+n) is not 0, for anyk,m, n. Conclude that θ is injective; and so BS(1, 2) ∼= A.

(h) Find the image of 1 under θ(a−kb2m+1ak+n). Hence show that no two elementsof L represent the same group element. That is, θ maps L bijectively to A.

(i) Give yourself a pat on the back: you’ve done some serious geometric grouptheory.

(j) Match the names “Gilbert Baumslag”, “Graham Higman” and “Donald Solitar”to their pictures below. Who are the others and what are they doing here?Could one of them be Johnny Winter, Max Dehn, Reidun Twarock or WilhelmMagnus?

(A) (B) (C)

(D) (E) (F)

(G) (H)

AJD January 2013

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MAS3219/MAS8219Groups,Graphs andSymmetry Assignment ...najd2/teaching/mas3219/assign.pdf · MAS3219/MAS8219Groups,Graphs andSymmetry Assignment Exercises: Solutions 1 Exercises forMonday