MAS8219 Groups, Graphs and Symmetry. Assignment Exercises: Solutions

MAS3219/MAS8219 Groups, Graphs and Symmetry Assignment Exercises: Solutions 1 Exercises for Monday 4th February This assignment will be assessed at ...
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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 one of the solutions to the class, on the board, during the problem class. Marks will be awarded as follows. For attendance: 1, for speaking at the board: 1, for making sense: 1, for a star performance: 1. Feedback on written solutions that you hand in at the problem class will be provided. 1.1 Fill in the details of the 2 statements following Definition 1.23 in the notes. Include the following. (a) Suppose that I is a (possibly infinite) set and that, for all i ∈ I, we have a subgroup Hi of a group G. Then H = ∩i∈I Hi is a subgroup of G. Using this fact explain (quickly) why h S i in Definition 1.23 is a group. From now on the question assumes the set up of Definition 1.23. (b) Explain (also quickly) why S ⊆ h S i.

(c) Explain (quickly again) why a subgroup containing S must also contain h S i.

(d) Use the above to show that h S i 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 (justifying any claims you make). (f) Let T = {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 = h S i.

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

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Suppose x, y ∈ H. Then, for all i ∈ I, we have x, y ∈ Hi , so from Lemma 1.4 we have xy −1 ∈ Hi . Therefore xy −1 ∈ ∩i∈I Hi = H. From Lemma 1.4 again, H is a subgroup of G. As h S i is an intersection of subgroups of G it now follows that h S i is also a subgroup of G. (b) By definition ∩{H ≤ G : H ⊇ S} = {x ∈ G : x ∈ H, for all H ≤ G such that H ⊇ S} and we have h S i = ∩{H ≤ G : H ⊇ S}.

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

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

(d) The “smallest subgroup containing S” is something like a least upper bound of subgroups 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 minimal subgroups 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 h S i 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, and is equal to h S i (and this holds for any subset S of G). We can now say “the smallest subgroup containing S”, instead of “a minimal subgroup containing S”. (e) Consider the element sn−εn · · · s1−ε1 of G. We have ε

−ε

n−1 (sε11 · · · sεnn )(sn−εn · · · s1−ε1 ) = (sε11 · · · sn−1 )sεnn sn−εn (sn−1n−1 · · · s1−ε1 )

ε

.. .

−ε

n−1 = (sε11 · · · sn−1 )(sn−1n−1 · · · s1−ε1 )

= sε11 s1−ε1 = 1G . (Strictly speaking this is an induction argument.) To see that (sn−εn · · · s1−ε1 )(sε11 · · · sεnn ) = 1 AJD January 2013

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we can replace each εi by −εi and relabel. Thus (sε11 · · · sεnn )−1 = sn−εn · · · s1−ε1 . (f) When n = 0 the sequence of length 0 is, by convention, the identity element of G. 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 1 n y −1 = t−δ · · · t−δ 1 . n

Thus 1 n xy −1 = sε11 · · · sεmm t−δ · · · t−δ 1 , n

which is again an element of T . That is, xy −1 ∈ T . Hence T is a subgroup of G. (g) By definition every element of s belongs to T ; so S ⊆ T . Moreover from the previous part of the question, T is a subgroup of G. Hence h S i ⊆ T . Every element of the form sε11 · · · sεmm , with si ∈ S and εi = ±1, belongs to h S i, as h S i is a group, so T ⊆ h S i. Hence T = h S i. 1.2 (a) Give the definitions of the order of a group and the order of an element of a group. (b) A group is cyclic if it can be generated by a single element. Show that if g belongs to the group G and has order n then {g} generates a cyclic subgroup of G of order n.

Solution (a) The order of an element g of a group G is the least positive integer such that g n = 1G . (b) The group h g i generated by {g}, is cyclic by definition. From the previous question it consists of elements of the form tε1 · P · · tεr , for some r ≥ 0 and εi = ±1. We may simplify tε1 · · · tεr to tm , where m = ri=1 εi , so h g i = {g m : m ∈ Z}. Let |g| = n, so n ≥ 1. For all m ∈ Z, we may write m = nq + r, where 0 ≤ r < n. Hence we have g m = g nq+r = (g n )q g r = g r . Thus any element of h g i may be written as g r , where 0 ≤ r < n. There are at most n such elements, so | h g i | ≤ n. Finally we must show that no two elements g r and g s , with r 6= s, are equal. Assume that 0 ≤ r < s < n. Then g r = g s if and only if g s−r = 1G , which implies that |g| ≤ s − r. As s − r < n this is contrary to the assumption that |g| = n. AJD January 2013

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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, given in Definition 2.1, is associative. (4 marks)

Solution. As (k, h)(k ′ , h′ ) = (k ◦ k ′ , hh′ ) we have, for k, k ′ , k ′′ ∈ K and h, h′ , h′′ ∈ H, ((k, h)(k ′ , h′ ))(k ′′ , h′′ ) = (k ◦ k ′ , hh′ )(k ′′ , h′′ ) = ((k ◦ k ′ ) ◦ k ′′ , (hh′ )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 Axiom 3 holds in K and H). (4 marks) 2.2 This question considers a couple of generating sets for the direct sum of an infinite cyclic group and the symmetric group S3 of degree 3. First we establish a correspondence between two descriptions of the infinite cyclic group. (a) Let C∞ denote the group with underlying set {xk : k ∈ Z} and binary operation xm xn = 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 isomorphism. (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. AJD January 2013

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i. The operation given is a binary operation. The axioms for a group must be checked. Axiom 1. We have x0 ∈ C∞ and, for all xm ∈ C∞ , x0 xm = xm = xm x0 , so x0 = 1C∞ . Axiom 2. If xm ∈ C∞ then x−m ∈ C∞ and xm x−m = x0 = x−m xm , so x−m = (xm )−1 . Hence Axiom 2 holds. Axiom 3. If xa , xb , xc ∈ C∞ then (xa xb )xc = xa+b xc = xa+b+c = xa xb+c = xa (xb xc ). Hence Axiom 3 holds. (3 marks) ii. If m, n ∈ Z then f (m + n) = xm+n = xm xn = 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, σ 2 , τ, στ, σ 2 τ }.

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 . We have immediately (1, σ), (0, τ ), (0, e) = (1, σ)0 and (−1, σ 2 ) = (1, σ)−1 in A. Then (0, τ )(1, σ) = (1, τ σ) = (1, σ 2 τ ) ∈ A, so (1, σ 2 τ )(0, τ ) = (1, σ 2 ) ∈ A. Therefore (−1, σ 2 )(1, σ 2 ) = (0, σ) ∈ A. Now we have the elements (0, σ 2) = (0, σ)2, (0, στ ) = (0, σ)(0, τ ) and (0, σ 2 τ ) = (0, σ 2 )(0, τ ) AJD January 2013

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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, σ m ) ∈ A, and (0, σ −m ) = (0, σ r ) ∈ A, where 0 ≤ r < 3 and −m ≡ r (mod 3). Thus (m, e) = (m, σ m )(0, σ −m) ∈ A. Thus (m, e) ∈ A, for all m ∈ Z. Hence (m, ρ) = (m, e)(0, ρ) ∈ A, for all m ∈ Z, ρ ∈ S3 . (5 marks) ii. Now let B be the subgroup of C∞ ⊕ S3 generated by (3, σ) and (2, τ ). Then B 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 , so B = 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} is a subgroup of Q. (d) Show that Z is a subgroup of Qp . (We know Z is a subgroup of Q, so it’s enough to 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 Q has 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 a non-trivial element of Q, written in this form, so r 6= 0. The order of r/s is the least positive integer n such that n(r/s) = 0, if such an integer exists, and is infinite 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. AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

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(b) Let r1 /s1 , . . . , rn /sn be elements of Q, with si > 0, and let H = hr1 /s1 , . . . rn /sn i be the subgroup they generate. We may assume that ri /si 6= 0, as the removal of 0 from the generating set of a subgroup leaves the subgroup unchanged. Let s = lcm(s1 · · · sn ) (any positive common multiple of s1 , . . . , sn will do) and let sˆi = s/si , for i = 1, . . . , n. Then s = si sˆi , so ri /si = ri sˆi /si sˆi = ri sˆi /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 (r1 sˆ1 /s) + · · · + mn (rn sˆn /s) = (m1 r1 sˆ1 + · · · + mn rn sˆn )/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 such that 1/(s + 1) = a/s, so s = a(s + 1). This implies s + 1 divides s, which in turn implies s + 1 ≤ s, since s and s + 1 are positive. This contradiction shows that 1/(s + 1) ∈ / H. We conclude that if H is a finitely generated subgroup of Q then there exists an element of Q not contained in H. This means that Q is not finitely generated. (6 marks) (c) The element 0/p ∈ Qp , so Qp 6= ∅. (Any element would do.) Let a/pk and b/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 a subgroup of Q. (2 marks) (d) For all m ∈ Z we have m = mp/p ∈ Qp , so Z ⊆ Qp . As Z is a subgroup of Q 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 Pr¨ ufer group or p-quasicyclic group.) AJD January 2013

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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 question by 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 k

k

xp = (exp(2πia/pk ))p = 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 prime this means d is a power of p, as required. [The following is a standard result in group theory. If an element g of a group satisfies g n = 1 then the order of the element g divides n. To see this, suppose the order of g is d > 0. Then we may write n = dq + r, where 0 ≤ r < d. Thus 1 = g n = g dq+r = g dq g r = (g d)q g r = g r , as g d = 1. Hence g r = 1, and as d is the smallest positive integer such that g d = 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 two cases. By definition a single element of Z(p∞ ) generates a cyclic group, and this is the base case for the induction. Now consider the subgroup generated by two non-trivial elements x = exp(2πia/pk ) and y = exp(2πib/pl ) of Z(p∞ ). Without loss of generality we may assume k ≤ l (otherwise swap x and y). Moreover, we may assume that p ∤ a and p ∤ b. Every element t of hx, yi has the form xm y n ,

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for some m, n ∈ Z, as Z(p∞ ) is an Abelian group. Therefore, if t ∈ hx, yi 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 1 number theory, S = {qd ∈ Z : d = gcd(apl−k , b), q ∈ Z}. Hence we have shown that if t ∈ hx, yi then t = exp(2πis/pl ), where s = qd, for some q ∈ Z, and d = 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 have t = xm y n , so t ∈ hx, yi. Hence hx, yi = {exp(2πidq/pl ) : q ∈ Z} = hzi, where z = 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 generated by k elements is cyclic. Let H be generated by elements x1 , . . . , xk+1 of Z(p∞ ). Then, from the inductive assumption, H ′ = hx1 , . . . , xk i is a cyclic group, say H ′ = hui, for some u ∈ Z(p∞ ). Moreover H = hH ′ , xk+1 i = hu, xk+1i, and so by the argument above, for two elements, H is also cyclic. As all elements of Z(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 = hzi, for some z = exp(2πid/pn ) ∈ Z(p∞ ). Consider the element v = exp(2πi/pn+1 ) of Z(p∞ ). If v ∈ H then v = z m , 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+1 q, for some q ∈ Z, and we obtain 1 = p(pn q + dm), so p|1, a contradiction. This shows that v ∈ / H, ∞ and so no finitely generated subgroup contains all elements of Z(p ). Hence Z(p∞ ) is not finitely generated. AJD January 2013

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(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 if m = pn q, for some q ∈ Z, if and only if m/pn ∈ Z. Therefore ker(θ) = Z. Thus Z(p∞ ) ∼ = Qp /Z, as claimed. (6 marks)

AJD January 2013

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Exercises for Friday 1st March

3.1 Let G be a group with subgroups H and K such that G is generated by the set H ∪ K. Show that if k −1 hk ∈ H, for all h ∈ H and k ∈ K, then H ⊳ G. (4 marks)

Solution Let h1 , h2 ∈ H. Then h−1 1 h2 h1 ∈ H, so for all h ∈ H and x ∈ H ∪ K we −1 have x hx ∈ H.

If g is an arbitrary element of G then g = xε11 · · · xεnn , for some xi ∈ H ∪ K and εi = ±1. As xi ∈ H ∪ K if and only if x−1 ∈ H ∪ K we can assume εi = 1, for all i i. Therefore, for all g ∈ G we have g = x1 · · · xn , with xi ∈ H ∪ K, for some n ≥ 0. Then −1 −1 −1 −1 g −1 hg = x−1 n · · · x1 hx1 · · · xn = xn · · · x2 h1 x2 · · · xn = · · · = xn hn−1 xn = hn ∈ H, −1 where h1 = x−1 1 hx1 and hi = xi hi−1 xi ∈ 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 remarks following 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 n1 k1 and n2 k2 belong to NK, with ni ∈ N and ki ∈ K, then (n1 k1 )(n2 k2 )−1 ∈ NK. We have −1 −1 −1 −1 −1 (n1 k1 )(n2 k2 )−1 = n1 k1 k2−1 n−1 2 = n1 (k1 k2 )n2 (k1 k2 ) (k1 k2 ) = n3 k3 , −1 −1 where n3 = (k1 k2−1 )n−1 ∈ N, as N ⊳ G, and k3 = k1 k2−1 ∈ K. Thus 2 (k1 k2 ) (n1 k1 )(n2 k2 )−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 of elements of the form sε11 · · · sεnn , where si ∈ N ∪ K and εi = ±1. Every element of NK is of this form, so NK ≤ H. However N ⊆ NK and K ⊆ NK, as elements n of N can be written as n · 1, with 1 ∈ K and elements k of K as 1 · k. Hence N ∪ K ⊆ NK. As H is the smallest subgroup containing N ∪ K we must then have H ≤ NK. Hence H = NK. 3.3 Let Γ be the graph shown below. ∗

0

00

1

01

10

11

This question will determine the structure of Sym(Γ), the group of isomorphisms of Γ. By considering the degrees of vertices we can see that every isomorphism of Γ maps ∗ to ∗, maps {0, 1} to {0, 1} and permutes the leaves of the graph. First it’s necessary to set up some notation. For a start, whenever describing the effect of an isomorphism 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 τ L the isomorphism of Γ such that τ L (0) = 0,

τ L (1) = 1,

τ L (00) = 01,

τ L (01) = 00,

τ L (1j) = 1j.

That is, τ L fixes all vertices except those below 0, which are swapped. Similarly, let τ R be the isomorphism of Γ such that τ R (0) = 0,

τ R (1) = 1,

τ R (0j) = 0j,

τ R (10) = 11,

τ R (11) = 10.

Then σ, τ L and τ R are isomorphisms of Γ. As (τ L )2 = (τ R )2 = Id it is also true that both L = {Id, τ L } and R = {Id, τ R } 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 vertices, so to check that two of them are the same it’s sufficient to check their effects on vertices.) AJD January 2013

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(a) Show that if ρ1 ∈ L and ρ2 ∈ R then ρ1 ρ2 = ρ2 ρ1 . [Consider the effect of the lhs and rhs on vertices.] Use this to show that LR = {ρ1 ρ2 : 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 sum of L and R and then, by quoting an appropriate theorem, that LR ∼ = L ⊕ R. (2 marks) (c) Show that the subgroup T = hσi 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 of LR and T . [It’s sufficient to show that if x ∈ T and y ∈ LR then x−1 yx ∈ LR. Consider the effect of x−1 yx 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 that S∼ = (LR) ⋊ T . (2 marks) (g) Explain why Sym(Γ) = S and hence that Sym(Γ) ∼ = (L ⊕ R) ⋊ T . [To see that every element of Sym(Γ) belongs to S consider the effect of an element on vertices. Break the argument into two cases. First suppose your element fixes 0. Then consider what happens if it swaps 0 and 1: try composing it with σ.] (7 marks)

Solution (a) If ρ1 = Id then ρ1 ρ2 = ρ2 = ρ2 ρ1 , as required. Similarly, if ρ2 = Id then the result holds. If ρ1 = τ L and ρ2 = τ R then ρ1 ρ2 (j) = ρ1 (j) = j = ρ2 (j) = ρ2 ρ1 (j), AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

14

for j ∈ {0, 1}, and, for x, y ∈ {0, 1}, ρ1 ρ2 (xy) = ρ1 (x[1 − y]) = [1 − x][1 − y], while ρ1 ρ2 (xy) = ρ1 ([1 − x]y) = [1 − x][1 − y]. Hence ρ1 ρ2 = ρ1 ρ2 , for all ρ1 ∈ L and ρ2 ∈ R. If f1 g1 and f2 g2 are elements of LR, where fi ∈ L and gi ∈ R, then (f1 g1 )(f2 g2 )−1 = f1 g1 g2−1f2−1 = f1 f2−1 g1 g2−1 ∈ LR, where the second equality holds because elements of L commute with elements of R, as we have just shown. From Lemma 1.4, as LR is not empty it is a subgroup of Sym(G). (4 marks) (b) We must check the conditions of Definition 2.3. We have just verified condition 3. Condition 1 holds by definition of LR. All that remains to be done is to check that L ∩ R = {Id}. If f ∈ L then f (1j) = 1j. As τ R (10) = 11 it follows that if f is also in R then f = Id. Hence L ∩ R = {Id}, and Theorem 2.6 implies that LR ∼ = L ⊕ R. (2 marks) (c) We must show that σ 2 = Id. As σ fixes all vertices except those of the form ij, i, j ∈ {0, 1}, we need only check the effect of σ 2 on such vertices. We have σ 2 (0j) = σ(1j) = 0j and σ 2 (1j) = σ(0j) = 1j. Hence σ 2 = Id and hσi = {Id, σ}, a cyclic group of order 2. (2 marks) (d) Let x ∈ T and y ∈ LR. If x = Id then certainly x−1 yx = y ∈ LR. Hence we may assume x = σ. As LR is the internal direct sum of L and R its elements are Id, τ L , τ R and τ L τ R . We have σ −1 = σ, as σ 2 = Id, so σ −1 τ L σ(0) = σ −1 τ L (1) = σ −1 (1) = 0 and σ −1 τ L σ(1) = σ −1 τ L (0) = σ −1 (0) = 1. Similarly σ −1 τ R σ(0) = 0 and σ −1 τ R σ(1) = 1. For y ∈ {0, 1}, σ −1 τ L σ(0y) = σ −1 τ L (1y) = σ −1 (1y) = 0y, and σ −1 τ L σ(1y) = σ −1 τ L (0y) = σ −1 (0[1 − y]) = 1[1 − y], AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

15

hence σ −1 τ L σ = τ R . We deduce that σ −1 τ R σ = σ −1 (σ −1 τ L σ)σ = τ L and then that σ −1 τ L τ R σ = σ −1 τ L σσ −1 τ R σ = τ R τ L = τ L τ R . Therefore, for all y ∈ LR, we have σ −1 yσ ∈ LR, and so LR is normal in the subgroup S. (8 marks) (e) As σ(0) = 1 and f (0) = 0, for all f ∈ LR, the isomorphism σ ∈ / LR. Hence LR ∩ T = {Id}. (1 mark) (f) We have shown that LR ⊳ S and that LR ∩ T = {Id}. To apply Theorem 4.11 we need to check that S = (LR)T . This follows from the result of Question 3.2(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 an isomorphism 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 does nothing 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 . Then f ′ ∈ Sym(G) and f ′ (0) = 0. Hence f ′ ∈ LR, using the result of the previous paragraph. Therefore f ∈ T (LR) = (LR)T = S. Thus, in all cases f ∈ S, so Sym(G) = S. (7 marks)

AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions 4

16

Exercises for Friday 15th March

4.1 Show that the isometry of R2 given by f (x, y) =

3−



! √ √ √ 3y 3x y x 3− − , −3 − 3 + − 2 2 2 2

is a direct isometry and a rotation and express it in the form (v, A) for appropriate v and A. Find its centre and angle of rotation. (10 marks)

Solution

so

 √  √ f (0, 0) = 3 − 3, −3 − 3 ,

√  3 − √3 v= . −3 − 3 ! √ √ 3y 3x y x , f (x, y) − f (0, 0) = − − , − 2 2 2 2 

so Ax =

− x2 − √

3x 2

and



3y 2



A=

y 2

!

=

− 12 − √ 3 2

− 12 − √

3 2



3 2

− 21



3 2

− 21 !

!

! x y

. (3 marks)

Thus f = (v, A). As det(A) =

1 4

+

3 4

= 1, f is direct, and as A 6= I2 , f is a rotation. (2 marks)

AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

c = (I2 − A)−1 v =

=

9 4

=

1 = 2

3 4

3 2 √ 3 2

1 2 1 √ 2 3

−1 √ 2 3 1 2

1 +

17 3 2 3 2

− 3 2

√ ! − 3 2 3 2

!



3 2 √

1 v= 2

!−1

v

v

1 √1 3

− √13

!

v

2

!

.

! √ √ 3− 3+ 3+1 = √ √ 3−1−3− 3

1

−2

Thus c = (2, −2)t . (4 marks) [Check: f (2, −2) = (2, −2), so this really is the centre of rotation.]

where θ =

2π , 3

  cos(θ) − sin(θ) , A= sin(θ) cos(θ) so f is rotation through

2π 3

radians, anticlockwise about (2, −2). (1 mark)

4.2 Show that the isometry (v, B) of R2 , given by   4x 3y 3x 4y g(x, y) = − + 2, − − +1 5 5 5 5 is an opposite isometry and a glide reflection and express it in the form (2a+b, B) for appropriate a, b and B. Find the equation of its axis of reflection and the distance of its minimal translation parallel to this axis. (12 marks)

Solution g(0, 0) = (2, 1)

AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions so

18

  2 v= 1

and

 4x 3y 3x 4y − ,− − g(x, y) − g(0, 0) = 5 5 5 5 ! ! ! 3y 4x 3 4 x − − 5 5 5 5 Bx = = 4y 3 4 3x y −5 −5 −5 − 5 

so

and

4 5

B=

− 53

− 35 − 54

!

−3

4

1 = 5

−3 −4

!

. (3 marks)

Bv =

1 (−16 25

− 9) = −1, so g is opposite. ! ! ! ! 4 −3 2 5 1 1 = 6= −v, = 5 −10 −3 −4 1 −2

Thus g = (v, B) and det(B) = 1 5

so g is a glide reflection. (2 marks) 2

1 1 a = (v − Bv) = 4 4 and b = v − 2a =

1 2 1

!

!

− 1 2 3 2



!

1 −2 =

!!

1 4 3 4

=

3 2

− 21

!

!

. (3 marks)

[Check: B should satisfy Ba = −a and Bb = b: ! ! 1! 5 − 1 4 −3 1 4 4 Ba = = = 3 15 5 −3 −4 5 − 4 4 and 1 Bb = 5

4

−3

−3 −4

!

3 2

− 21

!

1 = 5

15 2 − 52

− 41

− 34 !

!

= −a,

= b.

AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

19

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 through the point a. The line through the origin in direction of b is the set of points ! 3 sb = s

that is s

2

, for s ∈ R,

− 12

3 −1

!

,

s ∈ R.

Therefore the axis of reflection consists of points ! ! ! 1 1 3 3s + 4 4 sb + a = s + 3 = , 3 −s + −1 4 4

s ∈ R.

These are points (x, y)t with 1 4 3 y = −s + 4

x = 3s +

which is equivalent to 1 = 3s 4 3 y − = −s 4

x−

or

1 − 3

  1 3 x− =y− . 4 4

Rearranging this we find that the axis of reflection is the line m with equation x 5 y=− + . 3 6 [Alternatively apply the standard formula: the line parallel to b and passing through a has equation b0 (y −a1 ) = b1 (x−a0 ), that is 23 (y − 34 ) = − 12 (x− 41 ). After rearranging this gives the same answer.] q q , so b is a translation parallel to m The vector b has length ||b|| = 94 + 41 = 10 4 q through a distance of 52 . AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

20 (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, respectively. Assume that L is rectangular so L = {αa + βb | α, β ∈ Z}, for fixed vectors a = (x, 0)t and b = (0, y)t, where we may assume 0 < x < y in R, and that O = {I, −I, B0 , Bπ }, where I is the identity matrix (so −I is the matrix of rotation 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 y axes, respectively. As g is reflection in a line parallel to the x-axis it is of the form (v, B0), for some vector v such that B0 v = −v. Assume that v = αa + βb, for some α, β ∈ R. Then −v = B0 v = αB0 a + βB0 b = αa − βb if and only if αa − βb = −αa − βb

if and only if α = 0. Thus (v, B0 ) is a reflection if and only if v = βb, for some β ∈ 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 the y-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 B0 βb = −βb) with axis parallel to the x-axis and passing through the point β2 b. That is, W contains a reflection with axis parallel to the x-axis and passing through the point (β/2)b, for all β ∈ Z.

(L × {I})σ also contains elements (αa + βb, B0 ) with α 6= 0, and these are glide reflections with axis parallel to the x-axis, passing through the point β2 b, and with translation through a distance of α in the direction of the x-axis.

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MAS3219/8219 Assignment Exercises: Solutions

21

(b) Similarly, by considering (L × {I})τ , show that W contains a reflection with axis 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 other than 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 rotation ρ = (0, −I). (1 mark) (e) By considering (L × {I})ρ, show that W contains a rotation through π with centre 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 W as 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 of reflection and centres of rotation, and marking one instance of each distinct type 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 some vector u such that Bπ u = −u. If u = αa + βb then Bπ u = αBπ a + βBπ b = −αa + βb = −u AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

22

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 the y-axis and passing through the point α2 a. (2 marks) (c) (L × {I})τ also contains elements (αa + βb, Bπ ) with β 6= 0, and these are glide reflections with axis parallel to the y-axis, passing through the point α2 a, 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 a rotation through π about c = (I − (−I))−1 (αa + βb) 1 = I(αa + βb) 2 β α = a + b. 2 2 Thus W contains rotations through π about (αa + βb)/2, for all α, β ∈ Z. (4 marks) (f)

AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

23

b 2

2

2

2 a

• There are two orbits of horizontal axes of reflection: dotted-and-dashed-red and dashed-green. • There are two orbits of vertical axes of reflection: dotted-and-dashed-yellow and dashed-black. • There are four orbits of centres of rotation, all on axes of reflection. One representative 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

MAS3219/8219 Assignment Exercises: Solutions 5

24

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 {σ n , σ n τ : n ∈ Z}, where τ 2 = 1 and σ −1 τ = τ σ. Draw a diagram showing the vertices σ n and σ n τ , 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 use examples 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 does H. (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 not always hold. (You can use examples from the notes, if any are suitable.) (4 marks) 5.3 Let G be a group and S a generating set for G. Let Γ be the Cayley graph Γ(G, S) of G with respect to S. Let Sym+ (Γ) be the symmetry group of the labelled, directed graph Γ. From the notes we know that G acts faithfully (in fact freely) on Γ. Therefore we have an injective homomorphism α : G −→ Sym+ (Γ). In this question we shall show that α is surjective, so G ∼ = Sym+ (Γ). To do this suppose that f is an element of Sym+ (Γ). Then we shall show that f = α(g), for some g ∈ G. To keep things straight, for all g ∈ G, we shall denote by vg the vertex g of Γ and by αg the image AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions

25

α(g) of g in Sym+ (Γ). This notation means that for all g, h ∈ G the action of g on the vertex h is represented as αg (vh ) = vgh , instead of gh, and that the edges of Γ incident 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 map on 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 ), for s ∈ 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 is an 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 incident to 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). [You will need to show that αg−1 = αg−1 .] Complete the proof that G ∼ = Sym+ (Γ). (3 marks)

AJD January 2013

MAS3219/8219 Assignment Exercises: Solutions 6

26

Exercises for Friday 10th May

6.1 Apply the Stallings folding algorithm to find a free basis for each of the following subgroups of F (x, y, z). Show all steps of the algorithm and write out a free basis and the rank of the subgroup. (a) H1 = h yz, y −1z, z −1 y i.

(b) H2 = h xzx−2 , xyzx−1 z −1 x−1 , xyz −1 y −1 x−1 i. (c) H3 = h xyz, xzyz, xyxz, xyzy i.

6.2 (a) Show that the subgroup of F (x, y) generated by x−1 y −1 xy and x−1 y −2xy 2 has rank 2. (b) Generalise this to show that, for all n ≥ 1, the free group F (x, y) contains a free group of rank n, generated by {x−1 y −i xy i : 1 ≤ i ≤ n}. [Use an argument based 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−1 y −i xy i , for all i ≥ 1. Show, quoting and using an appropriate theorem, that there is a unique homomorphism 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 there is a positive integer n such that w is contained in the subgroup generated by z1 , . . . , 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) = ha, b | a−1bm a = bn i, for positive integers m, n are called Baumslag-Solitar groups (as they were first investigated by Graham Higman). Here we consider BS(1, 2), the group with presentation ha, b | a−1 ba = b2 i.

First we shall show that every element of BS(1, 2) can be written as an element of the set L = {an : n ∈ Z} ∪ {a−k b2m+1 ak+n : k, m, n ∈ Z}. This can be done by induction on the length of words in the generators. First it helps to establish the following fact for elements of BS(1, 2). (a) Show, by induction on r, that for all r ≥ 1, r

a−r bar = b2 in BS(1, 2).

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MAS3219/8219 Assignment Exercises: Solutions

27

Now for the proof that every element of BS(1, 2) can be represented by and element of L. The unique word of length 0 is the empty word, and this represents the identity element which can be written as a0 , so is in L. Assume that all words in F (a, b), of length at most n, are equal in BS(1, 2) to elements of L. To complete the inductive step 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 is represented by an element of L. We check each possible form for v, for each possible value of x. The possible forms of v are v = an or v = a−k b2m+1 ak+n . (b) Suppose that v = an . Check that a±1 v and b±1 v are elements of L. (No more than three lines required.) (c) Suppose that v = a−k b2m+1 ak+n and x = a±1 . Show (two lines only) that xv is in L. (d) Suppose that v = a−k b2m+1 ak+n , k ≥ 0 and x = b±1 . (The argument when k < 0 is given below.) Show that xv can be represented by an element of L, in the case where k ≥ 0. [Hint. Replace a−k b2m+1 ak+n by a power of b, using the first part of the question. It may help to remember that for all group elements g, h it’s true that g −1hk g = (g −1 hg)k .] (In part 3d above the case k < 0 was omitted. It works a little differently to the case k ≥ 0. (Thanks to David Robertson for spotting this). If v = a−k b2m+1 ak+n , where k < 0, then, for ε = ±1, bε v = bε a−k b2m+1 ak+n

= (a−k ak )bε a−k b2m+1 ak+n = a−k (ak ba−k )ε b2m+1 ak+n k

= a−k (b2 )ε b2m+1 ak+n ( as −k > 0) k ×ε+2m+1

= a−k b2

ak+n

k−1 ×ε+m)+1)

= a−k b2(2

ak+n

= a−k b2M +1 ak+n , where M = 2k−1 × ε + m ∈ Z. Hence bε v has the required form in this case too. )

This completes the inductive step, so now we have shown that every element of BS(1, 2) can be written as an element of L. Next we shall show that BS(1, 2) acts on the real line. Consider the two maps α and β of R to itself given by α(x) = x/2 and β(x) = x + 1. Together α and β generate a subgroup A of the group of invertible maps from R to R, under composition of maps: that is A = hα, βi. Define a map f : {a, b} −→ A by f (a) = α and f (b) = β. AJD January 2013

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28

(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 fact BS(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 is represented 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−k b2m+1 ak+n ) is not 0, for any k, m, n. Conclude that θ is injective; and so BS(1, 2) ∼ = A.

(h) Find the image of 1 under θ(a−k b2m+1 ak+n ). Hence show that no two elements of 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 group theory. (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 Wilhelm Magnus?

(A)

(B)

(D)

(C)

(E)

(G)

(F)

(H) AJD January 2013