SOLUTIONS OF HOMEWORK PROBLEMS

SOLUTIONS OF HOMEWORK PROBLEMS 31 If a 1, a,, a t1, a t are elements in a group G, prove that By (v) of Theorem 3 we have (a 1 a a t1 a t ) 1 a 1 t (a 1 a a t1 a t ) 1 a 1 t a 1 t1 a 1 a 1 1 (a 1 a a t1 ) 1 a 1 t a 1 t1(a 1 a a t ) 1 a 1 t a 1 t1 a 1 a 1 1 34 (i) How many elements of order are there in S 5 and in S 6? (ii) How many elements of order are there in S n? Answer: Case A: Let n k Then there are ( )( ) ( )( )( ) n n n n n 4 ( ) n! 3! elements of order in S n Case B: Let n k 1 Then there are ( )( ) ( )( )( ) n n n n n 4 ( ) n! 3! ( n ( n )( n k! )( n k! ) ) ( ) ( ) 3 elements of order in S n In particularly, it follows that S 4 has ( ) 4 ( 4 )( )! 9 elements of order and S 5 has elements of order ( ) 5 ( 5 )( ) 3! 5 35 If G is a group, prove that the only element g G with g g is 1 We rewrite g g as g g g Multiplying both sides by g 1, we get (g g)g 1 g g 1 g(g g 1 ) g g 1 g 1 1 g 1 1

36 Let H be a set containing an element e, and assume that there is an associative operation on H satisfying: 1 e x x for all x H; for every x H, there is x H with x x e (i) Prove that if h H satisfies h h h, then h e (ii) For all x H, prove that x x e (iii) For all x H, prove that x e x (iv) Prove that if e H satisfies e x x for all x H, then e e (v) Let x H Prove that if x H satisfies x x e, then x x (vi) Prove that H is a group (i) Multiplying both sides of h h h by h 1, we get h 1 (h h) h 1 h (h 1 h) h h 1 h e h e h e (ii) Consider (x x ) (x x ) We have (x x ) (x x ) x x (x x ) x (x x) x x (e x ) x x So, (x x ) (x x ) x x, and the result follows by (i) (iii) We have x (x x) x e On the other hand, by (ii) we get (x x ) x e x x Since x (x x) (x x ) x by the associative law, the result follows (iv) Since e x x for all x H, putting x e, we have On the other hand, by (iii) we get and the result follows (v) We have On the other hand, by (ii) and (iii) we get e e e e e e, (x x) x e x x x (x x ) x e x Since x (x x ) (x x) x by the associative law, the result follows (vi) Since there is an associative operation, e x x e and x x x x for any x H, it follows that H is a group

37 Let y be a group element of order m; if m tp for some prime p, prove that y t has order p Let g be the order of y t It follows that (y t ) g 1, so y tg 1 Note that g p, because otherwise g < p t tg < tp mtp tg < m So, y tg 1 and tg < m, which is impossible, since m is the smallest positive number with y m 1 Finally, we note that (y t ) p y tp y m 1 So, p is the smallest power of y t which gives 1, ie g p Remark: Note that the solution above does not use the fact that p is a prime In other words, problem 37 is true for any positive integer p 38 Let G be a group and let a G have order pk for some prime p, where k 1 Prove that if there is x G with x p a, then the order of x is p k Let g be the order of x We first note that therefore x pk (x p ) pk a pk 1, g p k We now show that p k is the smallest power which gives 1 In fact, since x g 1, we get therefore by Theorem 7 we obtain pk g, hence for some integer d So, (x g ) p 1 (x p ) g 1 a g 1, g pkd 1 x g x pkd (x p ) kd a kd But pk is the order of a, therefore kd pk, so d p This and ( ) give From ( ) and ( ) follows g p k ( ) ( ) g pkd p k ( ) 39 Let G GL(, Q), and let 0 1 A 1 0 and B 1 1 Show that A 4 E B 6, but that (AB) n E for all n > 0 We have A 1 0 0 1, A 4 1 0 0 1 1 0 3

and 1 1 1 1 0 1 B, B 4, 1 0 1 1 1 1 0 1 1 0 B 6 B 4 B 1 1 Finally, we show by induction that 1 n (AB) n In fact, for n 1 this is true, since 0 1 AB 1 0 1 1 1 1 Suppose this is true for some n k 1, that is 1 k (AB) k We prove that 1 k 1 (AB) k1 We have 1 k 1 1 1 k 1 (AB) k1 (AB) k (AB) 40(i) Prove, by induction on n 1, that n cos θ sin θ cos nθ sin nθ sin θ cos θ sin nθ cos nθ By n 1 this is obviously true Suppose this is true for some n k 1, that is cos θ sin θ sin θ cos θ k cos kθ sin kθ sin kθ cos kθ We prove that cos θ sin θ k1 cos(k 1)θ sin(k 1)θ sin θ cos θ sin(k 1)θ cos(k 1)θ 4

We have cos θ sin θ k1 cos θ sin θ k cos θ sin θ sin θ cos θ sin θ cos θ cos kθ sin kθ sin θ cos θ cos θ sin θ sin kθ cos kθ sin θ cos θ cos kθ cos θ sin kθ sin θ cos kθ sin θ sin kθ cos θ sin kθ cos θ cos kθ sin θ sin kθ sin θ cos kθ cos θ It is known that sin(α β) sin α cos β sin β cos α and cos(α β) cos α cos β sin α sin α, therefore and the result follows cos kθ cos θ sin kθ sin θ cos(kθ θ) cos(k 1)θ, sin kθ cos θ cos kθ sin θ sin(kθ θ) sin(k 1)θ, 41 If G is a group in which x 1 for every x G, prove that G must be abelian We have x x 1 for every x G This, in particularly, gives (ab)(ab) 1 for every a, b G Multiplying both sides by ba, we get ababba ba aba1a ba abaa ba ab1 ba ab ba, and the result follows 4 If G is a group with an even number of elements, prove that the number of elements in G of order is odd In particular, G must contain an element of order Let us split G into three subsets: G {e} {elements of order } {elements of order > } }{{}}{{} S 1 S Note that x 1 x x 1 5

This means that an element x coincides with its inverse if and only if x 1 In other words, an element x coincides with its inverse if and only if x has order or x 1 From this it follows that for any element x S we have x x 1 Moreover, it is easy to see that and x 1 x x 1 1 x 1 order of x order of x 1 The information above immediately implies that S has even number of elements Since G is even and {e} is odd, it follows that S 1 has odd number of elements 44 The stochastic group Σ(, R) consists of all those matrices in GL(, R) whose column sums are 1; that is, Σ(, R) consists of all the nonsingular matrices a c b d with a b 1 c d Prove that the product of two stochastic matrices is again stochastic, and that the inverse of a stochastic matrix is stochastic We have which is stochastic, since and Also, a c b d a1 a c 1 b a 1 c c 1 d b 1 a d 1 b b 1 c d 1 d a 1 a c 1 b b 1 a d 1 b a (a 1 b 1 ) b (c 1 d 1 ) a b 1 a 1 c c 1 d b 1 c d 1 d c (a 1 b 1 ) d (c 1 d 1 ) c d 1 a c which is stochastic, since d ad bc b d b ad bc 1 1 d c ad bc b a d b ad bc d b ad bd bd bc d ad bc b ad bc, c ad bc a ad bc d b d(a b) b(d c) d b d b 1 and c ad bc a ad bc a c ad bc a c ad ac ac bc a c a(d c) c(a b) a c a c 1 6

45 (i) Define a special linear group by Prove that SL (, R) is a subgroup of GL (, R) SL(, R) {A GL(, R) : det(a) 1} (ii) Prove that GL (, Q) is a subgroup of GL (, R) (i) We will use Theorem 4 In order to apply this Theorem we should check that SL (, R) is nonempty and that M 1 M 1 SL (, R) whenever M 1 SL (, R) and M SL (, R) 1 0 First of all note that SL (, R) is nonempty, since SL (, R) Let M 1 be from SL (, R) Then det(m 1 M 1 ) 1, since and M a c b d det(m 1 M 1 ) det(m 1 ) det(m 1 ) det(m 1 )det(m ) 1 1 1 1 1 Finally, M 1 M 1 a c b d 1 d a d b c b c a d b c a a d b c a d b c, which is equal to d c b a since a d b c det M 1 1 So, d c a1 d c 1 b a 1 c c 1 a M 1 M 1 b a b 1 d d 1 b b 1 c d 1 a, From this, obviously, follows that if M 1 and M have real entries, then M 1 M 1 has also real entries So, M 1 M 1 SL (, R) whenever M 1 SL (, R) and M SL (, R) (ii) To prove that GL (, Q) is a subgroup of GL (, R), we use Theorem 4 again First of all 1 0 note that GL (, Q) is nonempty, since GL (, Q) Let M 1 be from GL (, Q) Then det(m 1 M 1 ) 0, since and M a c b d det(m 1 M 1 ) det(m 1 ) det(m 1 ) det(m 1 )det(m ) 1, 7

which is nonzero, since det M 1 0 and det M 0 Finally, M 1 M 1 a c b d 1 d a d b c b c a d b c a a d b c a d b c a 1 d c 1 b a d b c a 1 c c 1 a a d b c b 1d d 1 b a d b c b 1 c d 1 a a d b c From this, obviously, follows that if M 1 and M have rational entries, then M 1 M 1 has also rational entries So, M 1 M 1 GL (, Q) whenever M 1 GL (, Q) and M GL (, Q) 46 Give an example of two subgroups H and K of a group G whose union H K is not a subgroup of G Let G V, H {(1), (1)(34)}, and K {(1), (13)(4)} Then H K {(1), (1)(34), (13)(4)}, which is not a subgroup, since it is not closed In fact, (1)(34)(13)(4) (14)(3) H K 48 If H and K are subgroups of a group G and if H and K are relatively prime, prove that H K {1} Let x be any element from H K First of all note that by Corollary we have a H 1 a K ( ) Let g be the order of x Then ( ) and Theorem 7 yield g H and g K Since H and K are relatively prime, it follows that g 1, which implies x 1 49 Let G a be a cyclic group of order n Show that a k is a generator of G if and only if (k, n) 1 ) Let G a be a cyclic group of order n and let a k be a generator of G We prove that (k, n) 1 First of all note that by Corollary we have a G 1 a n 1 We now suppose to the contrary that (k, n) l > 1 Then k ld 1 and n ld for some d 1, d Z Then (a k ) d (a ld 1 ) d (a ld ) d 1 (a n ) d 1 1 d 1 1 Note that since l > 1, we have d < n This means that the order of a k is < n This is impossible, because by Theorem 6 the order of a generator should be equal to the order of G, which is n ) Let G a be a cyclic group of order n and let (k, n) 1 We prove that a k is a generator of G In fact, let g be the order of a k Then (a k ) g 1 ( ) 8

On the one hand, since a n, by Theorem 6 the order of a is n From this and ( ) by Theorem 7 we get n kg This by Euclid s Lemma gives n g, since (k, n) 1 So, n g On the other hand, so g n Therefore n g (a k ) n (a n ) k 1 k 1, 50 Prove that every subgroup S of a cyclic group G a is itself cyclic If S {1}, we are done, since {1} is cyclic Suppose S {1}, that is S {1, a k 1, a k, } Let k be the smallest positive integer such that a k S If S a k, we are done Suppose S a k This means, that there exists a k i S such that k i dk Therefore by the Division Algorithm we get k i dk r, 1 r < k, so a k i a dkr a dk a r, hence a r a k i a dk Clearly, a k i, a dk S, therefore a r S, which is impossible, since 1 r < k and k is the smallest positive integer such that a k S 51 Prove that if G is a cyclic group of order n and if d n, then G has a subgroup of order d Let G a By Theorem 6, the order of a is n Then a n 1 Since d n, we have n kd Consider H a k and let g be the order of a k By Theorem 6 we have H g We prove that the order of H is d, ie g d In fact, since g is the order of a k, it follows that (a k ) g 1, so a kg 1 Note that g d, because otherwise g < d k kg < kd nkd kg < n So, a kg 1 and kg < n, which is impossible, since n is the smallest positive number with a n 1 Finally, we note that (a k ) d a kd a n 1 So, d is the smallest power of a k which gives 1, ie g d 5 Let G be a group of order 4 Prove that either G is cyclic or x 1 for every x G Conclude, using Exercise 41, that G must be abelian Pick some x G and consider H a If H 4, then H G, therefore G is cyclic and we are done Suppose H < 4 Since H is a subgroup of G, by Lagrange s Theorem we have H or H 1, therefore by Theorem 6 the order of x is (which means, that x 1) or 1 (which means, that x 1, so x 1 again) So, in both cases x 1 Finally, to show that G is abelian, we note that if G is cyclic, then it is abelian, since all cyclic groups are abelian If x 1 for every x G, then G is abelian by Exercise 41 9