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GROUP THEORY EXERCISES

AND SOLUTIONS

Mahmut Kuzucuo˘glu

Middle East Technical University

matmah@metu.edu.tr

Ankara, TURKEY

November 10, 2014

ii

iv

v

Preface

I have given some group theory courses in various years. These problems are given to students from the books which I have followed that year. I have kept the solutions of exercises which I solved for the students. These notes are collection of those solutions of exercises.

Mahmut Kuzucuo˘glu METU, Ankara November 10, 2014

GROUP THEORY EXERCISES AND SOLUTIONS

M. Kuzucuo˘glu

1. SEMIGROUPS

Definition A semigroup is a nonempty set S together with an associative binary operation on S. The operation is often called mul- tiplication and if x, y ∈ S the product of x and y (in that ordering) is written as xy.

1.1. Give an example of a semigroup without an identity element.

Solution Z+^ = { 1 , 2 , 3 , ...} is a semigroup without identity with binary operation usual addition.

1.2. Give an example of an infinite semigroup with an identity element e such that no element except e has an inverse.

Solution N= { 0 , 1 , 2 , ...} is a semigroup with binary operation usual addition. No non-identity element has an inverse.

1.3. Let S be a semigroup and let x ∈ S. Show that {x} forms a subgroup of S (of order 1) if and only if x^2 = x such an element x is called idempotent in S.

Solution Assume that {x} forms a subgroup. Then {x} ∼= { 1 } and x^2 = x. 1

2 M. KUZUCUO GLU˘

Conversely assume that x^2 = x. Then associativity is inherited from S. So Identity element of the set {x} is itself and inverse of x is also itself. Then {x} forms a subgroup of S.

2. GROUPS Let V be a vector space over the field F. The set of all linear in- vertible maps from V to V is called general linear group of V and denoted by GL(V ).

2.1. Suppose that F is a finite field with say |F | = pm^ = q and that V has finite dimension n over F. Then find the order of GL(V ).

Solution Let F be a finite field with say |F | = pm^ = q and that V has finite dimension n over F. Then |V | = qn^ for any base w 1 , w 2 , ..., wn of V , there is unique linear map θ : V → V such that viθ = wi for i = 1, 2 , ..., n. Hence |GL(V )| is equal to the number of ordered bases of V , in form- ing a base w 1 , w 2 , ..., wn of V we may first choose w 1 to be any nonzero vector of V then w 2 be any vector other than a scalar multiple of w 1. Then w 3 to be any vector other than a linear combination of w 1 and w 2 and so on. Hence |GL(V )| = (qn^ − 1)(qn^ − q)(qn^ − q^2 )....(qn^ − qn−^1 ).

2.2. Let G be the set of all matrices of the form

a b 0 c

where

a, b, c are real numbers such that ac 6 = 0. (a) Prove that G forms a subgroup of GL 2 (R).

Indeed (^) ( a b 0 c

d e 0 f

ad ae + bf 0 cf

∈ G

ac 6 = 0, df 6 = 0, implies that acdf 6 = 0 for all a, c, d, f ∈ R. Since determinant of the matrices are all non-zero they are clearly invertible. (b) The set H of all elements of G in which a = c = 1 forms a subgroup

of G isomorphic to R+. Indeed H = {

1 b 0 1

| b ∈ R }

4 M. KUZUCUO GLU˘

Solution φ(x 1 x 2 ) = [(x 1 x 2 )−^1 ]t = [x− 2 1 x− 1 1 ]t = (x− 1 1 )t(x− 2 1 )t^ = φ(x 1 )φ(x 2 )

Now if φ(x 1 ) = 1 = (x− 1 1 )t, then x− 1 1 = 1. Hence x 1 = 1. So φ is a monomorphism. For all x ∈ GLn(F ) there exists x 1 ∈ GLn(F ) such that φ(x 1 ) = x. Let x 1 = (x−^1 )t. So we obtain φ is an automorphism. Let H = {x ∈ GLn(F ) : φ(x) = x}. We show in the previous exercise that H is a subgroup of GLn(F ). Now for x ∈ H φ(x) = x = (x−^1 )t implies xxt^ = 1. That is the set of the orthogonal matrices.

Recall that if G = G 1 × G 2 , then the subgroup H of G may not be of the form H 1 × H 2 as H = {(0, 0), (1, 1)} is a subgroup of Z 2 × Z 2 but H is not of the form H 1 × H 2 where Hi is a subgroup of Gi. But the following question shows that if |G 1 | and |G 2 | are relatively prime, then every subgroup of G is of the form H 1 × H 2.

2.5. Let G = G 1 × G 2 be a finite group with gcd(|G 1 |, |G 2 |)) = 1. Then every subgroup H of G is of the form H = H 1 × H 2 where Hi is a subgroup of Gi for i = 1, 2.

Solution: Let H be a subgroup of G. Let πi be the natural projection from G to Gi. Then the restriction of πi to H gives homo- morphisms from H to Gi for i = 1, 2. Let Hi = πi(H) for i = 1, 2. Then clearly H ≤ H 1 × H 2 and Hi ≤ Gi for i = 1, 2. Then H/Ker(π 1 ) ∼= H 1 implies that |H 1 | | |H| similarly |H 2 | | |H|. But gcd(|H 1 |, |H 2 |) = 1 implies that |H 1 ||H 2 | | |H|. So H = H 1 × H 2.

2.6. Let H  G and K  G. Then H ∩ K  G. Show that we can define a map

ϕ : G/H ∩ K −→ G/H × G/K g(H ∩ K) −→ (gH, gK) for all g ∈ G and that ϕ is an injective homomorphism. Thus G/(H ∩ K) can be embedded in G/H × G/K. Deduce that if G/H and G/K or both abelian, then G/H ∩ K abelian.

GROUP THEORY EXERCISES AND SOLUTIONS 5 Solution As H and K are normal in G, clearly H ∩ K is normal in G. ϕ : G/H ∩ K −→ G/H × G/K

ϕ(g(H ∩ K)g′(H ∩ K)) = ϕ(gg′(H ∩ K)) = (gg′H, gg′K) = (gH, gK)(g′H, g′K) = ϕ(g(H ∩ K))ϕ(g′(H ∩ K)).

So ϕ is an homomorphism. Kerϕ = {g(H ∩ K) : ϕ(g(H ∩ K)) = (¯e, ¯e) = (gH, gK)}. Then g ∈ H and g ∈ K implies that g ∈ H ∩K. So Kerϕ = H ∩K. If G/H and G/K are abelian, then g 1 Hg 2 H = g 1 g 2 H = g 2 g 1 H. Similarly g 1 g 2 K = g 2 g 1 K for all g 1 , g 2 ∈ G, g− 2 1 g 1 − 1 g 2 g 1 ∈ H, g 2 − 1 g− 1 1 g 2 g 1 ∈ K. So for all g 1 , g 2 ∈ G, g− 2 1 g 1 − 1 g 2 g 1 ∈ H ∩ K. g 2 − 1 g− 1 1 g 2 g 1 (H ∩ K) = H ∩ K. So g 2 g 1 (H ∩ K) = g 1 g 2 (H ∩ K).

2.7. Let G be finite non-abelian group of order n with the property that G has a subgroup of order k for each positive integer k dividing n. Prove that G is not a simple group.

Solution Let |G| = n and p be the smallest prime dividing |G|. If G is a p-group, then 1 6 = Z(G)  G. Hence G is not simple. So we may assume that G has composite order. Then by assumption G has a subgroup M of index p in G. i.e. |G : M | = p. Then G acts on the right cosets of M by right multiplication. Hence there exists a homomorphism φ : G ↪→ Sym(p). Then G/Kerφ is isomorphic to a subgroup of Sym(p). Since p is the smallest prime dividing the order of G we obtain |G/Kerφ|| p! which implies that |G/Kerφ| = p. Hence Kerφ 6 = 1 otherwise Ker φ = 1 implies that G is abelian and isomorphic to Zp. But by assumption G is non-abelian.

2.8. Let M ≤ N be normal subgroups of a group G and H a subgroup of G such that [N, H] ≤ M and [M, H] = 1. Prove that for all h ∈ H and x ∈ N (i) [h, x] ∈ Z(M )

GROUP THEORY EXERCISES AND SOLUTIONS 7 2.9. Let G be a finite group and Φ(G) the intersection of all max- imal subgroups of G. Let N be an abelian minimal normal subgroup of G. Then N has a complement in G if and only if N 5 Φ(G)

Solution Assume that N has a complement H in G. Then G = N H and N ∩ H = 1. Since G is finite there exists a maximal subgroup M ≥ H. Then N is not in M which implies N is not in Φ(G). Because, if N ≤ M , then G = HN ≤ M which is a contradiction. Conversely assume that N  Φ(G). Then there exists a maximal subgroup M of G such that N  M. Then by maximality of M we have G = N M. Since N is abelian N normalizes N ∩ M hence G = N M ≤ NG(N ∩ M ) i.e. N ∩ M is an abelian normal subgroup of G. But minimality of N implies N ∩M = 1. Hence M is a complement of N in G.

2.10. Show that F (G/φ(G)) = F (G)/φ(G). Solution: (i) F (G)/φ(G) is nilpotent normal subgroup of G/φ(G) so F (G)/φ(G) ≤ F (G/φ(G)). Let K/φ(G) = F (G/φ(G)). Then K/φ(G) is maximal normal nilpotent subgroup of G/φ(G). In particular K E G and K/φ(G) is nilpotent. It follows that K is nilpotent in G. This implies that K ≤ F (G). K/φ(G) ≤ F (G)/φ(G) which implies F (G/φ(G)) = F (G)/φ(G).

2.11. If F (G) is a p-group, then F (G/F (G)) is a p′- group. Solution: Let K/F (G) = F (G/F (G)), maximal normal nilpotent subgroup of G/F (G). So K/F (G) = Dr Oq(K/F (G)) q∈Π(G)

= P 1 /F (G) ×

P 2 /F (G) ×... × Pm/F (G). Since F (G) is a p-group so one of Pi/F (G) is a p-group, say P 1 /F (G) is a p-group. Now P 1 is a p-group, P 1 /F (G)charK/F (G)charG/F (G) implies that P 1 /F (G)charG/F (G) implies P 1 / G. This implies P 1 is a p-group and hence nilpotent and normal implies P 1 ≤ F (G). So P 1 /F (G) = id i.e K/F (G) = F (G/F (G)) is a p′-group.

Observe this in the following example. S 3 , F (S 3 ) = A 3. F (S 3 /A 3 ) = S 3 /A 3 ∼= Z 2 is a 2-group.

2.12. Let G = {(aij ) ∈ GL(n, F ) | aij = 0 if i > j and aii = a, i = 1.... , n} where F is a field, be the group of upper triangular

8 M. KUZUCUO GLU˘

matrices all of whose diagonal entries are equal. Prove that G ∼= D × U where D is the group of all non-zero multiples of the identity matrix and U is the group of upper triangular matrices with 1’s down diagonal.

Solution

^ d:^ G^ →^ F^ ∗     

a c 12 c 13 c 14 ... c 1 n 0 a c 23 c 24 ... c 2 n .

. ... ∗ 0 0 0 0 a cn− 1 n 0 0 0 0 0 a

→ a

It is clear that d is a homomorphism and Ker d = U. So U is normal D ∩ U = 1. Since F is a field and a is a non-zero element every element g ∈ G can be written as a product g = cu where c ∈ D and u ∈ U. So DU = G. Moreover D is normal in G in fact D is central in G. So G = DU ∼= D × U.

2.13. Prove that if N is a normal subgroup of the finite group G and (|N |, |G : N |) = 1, then N is the unique subgroup of order |N |.

Solution If M is another subgroup of G of order |N |. Then N M is a subgroup of G as N
G. Now |N M | = ||NN^ ||∩MM^ ||. If N 6 = M , then |N M | > |N | and if π is the set of primes dividing |N |, then N is a maximal π-subgroup of G. But M N is also a π-group containing N properly. Hence M N = N. i.e M ≤ N.

2.14. Let F be a field. Define a binary operation ∗ on F by a ∗ b = a + b − ab for all a, b ∈ F. Prove that the set of all elements of F distinct from 1 forms a group F x^ = F \ { 1 } with respect to the operation ∗ and that F ∗^ ∼= F x^ where F ∗^ is the multiplicative group on F \ { 0 } with respect to the usual multiplication in the field.

10 M. KUZUCUO GLU˘

G^ ˆ ∼= G. Indeed define

ϕ : G −→ Gˆ g −→ (g, g)

ϕ(gg′) = (gg′, gg′) = (g, g)(g′, g′) = ϕ(g)ϕ(g′). So ϕ is a homomor- phism. ϕ(g) = 1 = (g, g). This implies g = 1. So ϕ is a monomorphism. For all (gi, gi) ∈ Gˆ there exists gi ∈ G such that ϕ(gi) = (gi, gi). So ϕ is onto. Hence ϕ is an isomorphism. ii) Gˆ  G × G if and only if G is abelian. Assume Gˆ is a normal subgroup of G × G. Then for any g 1 , g 2 ∈ G, (g 1 , g 2 )−^1 (x, x)(g 1 , g 2 ) = (g− 1 1 xg 1 , g 2 − 1 xg 2 ) ∈ Gˆ. In particular g 1 = 1 implies for all g 2 , and for all x ∈ G, g 2 − 1 xg 2 = x. Hence G is abelian. Conversely if G is abelian, then G×G is abelian and every subgroup of G × G is normal in G, in particular Gˆ is normal in G.

2.16. Suppose H  G. Show that if x, y elements in G such that xy ∈ H, then yx ∈ H.

Solution H  G, implies that every left coset is also a right coset Hx = xH, yH = Hy, xy ∈ H so H = xyH. xH = Hx implies xyxH = xyHx = Hx. Then yxH = x−^1 Hx = H. Hence yx ∈ H.

2.17. Give an example of a group such that normality is not tran- sitive.

Solution Let us consider A 4 alternating group on four letters. Then V = { 1 , (12)(34), (13)(24), (14)(23)} is a normal subgroup of A 4. Since V is abelian any subgroup of V is a normal subgroup of V. But H = { 1 , (12)(34)} is not normal in A 4.

Another Solution Let’s consider G = S 3 ×S 3 , A 3 = { 1 , (123), (132)}. A 3 / S 3. Let A = { (1, 1), ((123), (123)), ((132), (132)) } ≤ G, A is diagonal subgroup of A 3 × A 3 and A ∼= A 3. A / A 3 × A 3 / G. But A is not normal in G as ((12), 1)−^1 ((123), (123))((12), 1) = ((132), (123)) ∈/ A.

GROUP THEORY EXERCISES AND SOLUTIONS 11 2.18. If α ∈ AutG and x ∈ G, then |xα| = |x|. Solution First observe that (xα)n^ = (xn)α. If xα^ has finite order say n, then (xα)n^ = 1 = (xn)α^ = 1α. Hence xn^ = 1 as α is an automor- phism. Hence x has finite order dividing n. If order of x is less than or equal to n, say m. Then we obtain xm^ = 1. Then (xm)α^ = 1α^ = 1. Hence (xα)m^ = 1. It follows that n = m, i.e. |xα| = |x| when the order is finite. But the above proof shows that if order of xα^ is infinite then order of x must be infinite. In particular conjugate elements of a group have the same order. We can consider the semidirect product of G with the Aut(G). Then in the semidirect product the elements x and xα^ becomes conjugate elements.

2.19. Let H and K be subgroups of G and x, y ∈ G with Hx = Ky. Then show that H = K.

Solution Hx = Ky implies Hxy−^1 = K. As H is a subgroup, 1 ∈ H and so xy−^1 ∈ Hxy−^1 = K. Then yx−^1 ∈ K. It follows that K = Kyx−^1. Then K = Kxy−^1 = Kyx−^1 = H. Hence K = H.

2.20. Prove that if K is a normal subgroup of the group G, then Z(K) is a normal subgroup of G. Show by an example that Z(K) need not be contained in Z(G).

Solution: Let z ∈ Z(K), k ∈ K and g ∈ G. Then g−^1 zg ∈ K as K E G and (g−^1 zg)k(g−^1 z−^1 g)k−^1 = g−^1 z(gkg−^1 )z−^1 gk−^1 = g−^1 (gkg−^1 )zz−^1 gk−^1 = 1. Hence Z(K) E G.

Now as an example consider A 3 in S 3. Z(A 3 ) = A 3 but Z(S 3 ) = 1. 2.21. Let x, y ∈ G and let xy = z if z ∈ Z(G), then show that x and y commute.

Solution: xy = z ∈ Z(G) implies for all g ∈ G, (xy)g = g(xy). This is also true for x, hence (xy)x = x(xy). Now multiply both side by x−^1 , we obtain yx = xy. Then x and y are commute.

2.22. Let U T (3, F ) be the set of all matrices of the form   

1 a b 0 1 c 0 0 1

GROUP THEORY EXERCISES AND SOLUTIONS 13 ϕ is an isomorphism.

Now to see that U T (3, F )/Z(U T (3, F )) ∼= F +^ × F +. Let θ : U T (3, F )/Z(U T (3, F )) −→ F +^ × F +.  

1 a b 0 1 c 0 0 1

 Z −→ (a, c)

θ is well defined and, moreover θ is an isomorphism.

(iii) Now all we need to do is to compare the order of U T (3, pm) and the order of the Sylow p-subgroup of GL(3, pm). It is easy to see that |U T (3, pm)| = p^3 m. And |GL(3, pm)| = (p^3 m−1)(p^3 m−pm)(p^3 m−p^2 m) = p^3 m((p^3 m^ − 1)(p^2 m^ − 1)(pm^ − 1)). Hence p part are the same and we are done.

2.23. Let x ∈ G, D := {xg^ : g ∈ G} and Ui ≤ G for i= 1,2. Suppose that 〈D〉 = G and D ⊆ U 1 ∪ U 2. Then show that U 1 = G or U 2 = G.

Solution: Assume that U 1 6 = G. Then there exists g ∈ G such that xg^ ∈/ U 1 otherwise all conjugates of x is contained in U 1 and so D ⊆ U 1 which implies U 1 = G. Then xg^ ∈/ U 1 implies xg^ ∈ U 2 as D ⊆ U 1 ∪ U 2. Now for any u 1 ∈ U 1 , (xg)u^1 ∈/ U 1 otherwise xg^ will be in U 1 which is impossible. Then for any u 1 ∈ U 1 we obtain (xg)u^1 ∈ U 2. Now U 2 is a subgroup and xg^ ∈ U 2 so we have (xg)u^2 ∈ U 2 for all u 2 ∈ U 2. As 〈U 1 ∪ U 2 〉 = G we obtain (xg)t^ ∈ U 2 for all t ∈ G, i.e, D ⊆ U 2 this implies 〈D〉 ≤ U 2 but 〈D〉 = G ≤ U 2 which implies U 2 = G.

2.24. Let g 1 , g 2 ∈ G. Then show that |g 1 g 2 | = |g 2 g 1 |. Solution: We will show that if |g 1 g 2 | = k < ∞, then |g 2 g 1 | = k. Let |g 1 g 2 | = k. ( ︸g 1 g 2 )(g 1 g︷︷ 2 ) ....(g 1 g (^2) ︸)

k−times

= 1. Then multiplying from left

by g− 1 1 and from right by g− 2 1 we have ( ︸g 2 g 1 )(g 2 g︷︷ 1 )... (g 2 g 1 )︸

(k−1)−times

= g 1 −^1 g 2 −^1.

Now multiply from right first by g 2 and then g 1 , we obtain ( ︸g 2 g 1 )(g 2 g︷︷ 1 ) ...(g 2 g (^1) ︸)

k−times

= ((g 2 g 1 ))k^ = 1. It cannot be less than k since we

14 M. KUZUCUO GLU˘

may apply the above process and then reduce the order of (g 1 g 2 ) less than k.

2.25. Let H ≤ G, g 1 , g 2 ∈ G. Then Hg 1 = Hg 2 if and only if g 1 −^1 H = g 2 −^1 H.

Solution: (⇒) If Hg 1 = Hg 2 , then H = Hg 2 g 1 −^1 hence g 2 g 1 −^1 ∈ H. Then H is a subgroup implies (g 2 g 1 −^1 )−^1 ∈ H i.e. g 1 g 2 −^1 ∈ H. It follows that g 1 g 2 −^1 H = H. Hence g 2 −^1 H = g 1 −^1 H. (⇐) If g 1 −^1 H = g 2 −^1 H, then g 1 g 2 −^1 ∈ H by the same idea in the first part we have (g 1 g 2 −^1 )−^1 ∈ H, g 2 g 1 −^1 ∈ H i.e. Hg 2 g 1 −^1 = H. This implies Hg 1 = Hg 2.

2.26. Let H ≤ G, g ∈ G if |g| = n and gm^ ∈ H where n and m are co-prime integers. Then show that g ∈ H.

Solution: The integers m and n are co-prime so there exists a, b ∈ Z satisfying an + bm = 1. Then g = gan+bm^ = gangbm^ = (gn)a(gm)b^ = gmb^ ∈ H. As H is a subgroup of G, gm^ ∈ H implies gbm^ ∈ H and gna^ = 1. Hence gmb^ = g ∈ H.

2.27. Let g ∈ G with |g| = n 1 n 2 where n 1 , n 2 co-prime positive integers. Then there are elements g 1 , g 2 ∈ G such that g = g 1 g 2 = g 2 g 1 and |g 1 | = n 1 , |g 2 | = n 2.

Solution: As n 1 and n 2 are relatively prime integers, there exist a and b in Z such that an 1 + bn 2 = 1. Observe that a and b are also relatively prime in Z. Then g = g^1 = gan^1 +bn^2 = gan^1 gbn^2. Let g 1 = gbn^2 and g 2 = gan^1. Then g 1 n 1 = (gbn^2 )n^1 = 1, gn 2 2 = (gan^1 )n^2 = 1 g = g 1 g 2 = gan^1 +bn^2 = gbn^2 +an^1 = g 2 g 1. Indeed |g 1 | = n 1. If gm 1 = 1, then m|n 1 and g 1 m = gbn^2 m^ = 1. It follows that n 1 n 2 |bn 2 m. Then n 1 |bm but by above observation n 1 and b are relatively prime as an 1 +bn 2 = 1, so n 1 |m. It follows that n 1 = m. Similarly |g 2 | = n 2.

2.28. Let g 1 , g 2 ∈ G with |g 1 | = n 1 < ∞, |g 2 | = n 2 < ∞, if n 1 and n 2 are co-prime and g 1 and g 2 commute, then |g 1 g 2 | = n 1 n 2.

Solution: The elements g 1 and g 2 commute. Therefore (g 1 g 2 )n^1 n^2 = gn 1 1 n^2 gn 2 1 n^2 = (g 1 n 1 )n^2 (g 2 n 2 )n^1 = 1. Assume |g 1 g 2 | = m. Then (g 1 g 2 )m^ = gm 1 gm 2 = 1. Then m|n 1 n 2 and g 1 m = g− 2 m. (gm 1 )n^1 = (g 2 − m)n^1 = 1. Then n 2 |mn 1 but gcd(n 1 , n 2 ) = 1. We obtain