Zachary Scherr Math 370 HW 7 Solutions

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1 1 Book Problems b Solution: Let U 1 {u 1 u U} and let S U U 1. Then (U) is the set of all elements of G which are finite products of elements of S. We are told that for any u U and g G we have gug 1 U. This implies that gu 1 g 1 (gug 1 ) 1 U 1 as well. Thus we see that every s S satisfies gsg 1 S for every g G. Let x (U). Then there exists k 0 and s 1, s 2,..., s k S so that x s 1 s 2 s k. We must prove that for every g G we have gxg 1 (U). We prove this by induction on k. The case k 0 is trivial (x e in this case) and the case k 1 follows from our statement about S. Thus we may assume the statement is true for any element of (U) expressible as a product of k 1 terms in S. If x s 1 s 2 s k, then for any g G we have gxg 1 g(s 1 s 2 s k )g 1 (gs 1 s 2 s k 1 g 1 )(gs k g 1 ). The induction hypothesis ensures gs 1 g k 1 g 1 (U) while the case k 1 ensures gs k g 1 (U). Since (U) is closed under multiplication, we also get that gxg 1 (U). Thus we ve proved that for every x (U) and g G, gxg 1 (U) which implies that (U) is normal in G Solution: (a) In light of 2.7.4b, we only need to show that for every u U and g G we have gug 1 U. Of course any u U is of the form u xyx 1 y 1 for some x, y G so gug 1 g(xyx 1 y 1 )g 1 (gxg 1 )(gyg 1 )(gxg 1 ) 1 (gyg 1 ) 1 which shows that gug 1 U as well. (b) For any a, b G we have aba 1 b 1 (ab)(ba) 1 U G. Thus we have an equality of cosets G ab G ba. But of course in the group G/G this just says which means that G/G is abelian. G ag b G ab G ba G bg a (c) Let N G be a normal subgroup and suppose G/N is abelian. Then for every a, b G we have Nab NaNb NbNa Nba which says that (ab)(ba) 1 aba 1 b 1 N. Thus U N and since N is a subgroup of G containing U we must have G N. (d) Let ϕ: G G/G be the onto homomorphism ϕ(g) G g. We proved in class that there is a one-to-one correspondence between subgroups of G/G and subgroups of G which contain G, the kernel of ϕ. In our correspondence, we also showed that normal subgroups of G/G correspond to normal subgroups of G. The fact that G/G is abelian means that every one of its subgroups is normal, and hence by the correspondence, every subgroup of G containing G must be normal in G.

2 Solution: Consider the function ϕ: N NM/M given by ϕ(n) Mn. Notice that ϕ actually gives a function since N NM. It is easy to verify that ϕ is a homomorphism since ϕ(n 1 n 2 ) M(n 1 n 2 ) Mn 1 Mn 2 ϕ(n 1 )ϕ(n 2 ). We can also check that ϕ is onto. Let X NM/M. Then there exists x NM so that X Mx. Of course then there must be n N and m M so that x nm. We have X Mx M(nm) MnMm MnMe Mn so that X ϕ(n). If K is the kernel of ϕ then the isomorphism theorems tell us that N/K NM/M. By definition, K {n N Mn Me} but Mn Me if and only if n M. Thus K N M and hence N/N M NM/M Solution: We define a function ϕ: G R {0} by ϕ(τ ab ) a. Notice that this is actually a function since by assumption a 0. We claim that ϕ is a homomorphism onto R {0}, as a group under multiplication, with kernel N. It s clear that ϕ is onto. To see it s a homomorphism, note that for any x V and a, b, c, d R with ac 0 we have Thus τ ab τ cd (x) τ ab (cx + d) a(cx + d) + b acx + ad + b τ (ac)(ad+b) (x). ϕ(τ ab τ cd ) ϕ(τ (ac)(ad+b) ) ac ϕ(τ ab )ϕ(τ cd ). It is clear by definition that ker(ϕ) consists of all τ 1b for b R so ker(ϕ) N. Thus immediately we see that N is a normal subgroup of G and in fact by the isomorphism theorems G/N R {0} Solution: (a) The subgroup N (y) is trivially normal in G since it has index 2 and you proved on the exam practice problems that an index two subgroup is always normal. To prove this without resorting to machinery, notice that normality of N will follow from knowing that xn Nx. Now xn {x, xy, xy 2,..., xy n 1 } Page 2

3 while Nx {x, yx, y 2 x,..., y n 1 x} {x, xy n 1, xy n 2,..., xy} xn. (b) This is obvious. G/N is a group of order 2 so it must be cyclic, and up to isomorphism there is only one cyclic group of order 2. For practice we use the isomorphism theorems. Consider a function ϕ: G {1, 1} defined by ϕ(x i y j ) ( 1) i. We ve previously seen that Thus (x i y j )(x i y j ) x i+i y j( 1)i. ϕ((x i y j )(x i y j )) ϕ(x i+i y j( 1)i ) ( 1) i+i ( 1) i ( 1) i ϕ(x i y j )ϕ(x i y j ) so ϕ is a homomorphism. The map ϕ is trivially onto and it is clear that N ker(ϕ). Thus the isomorphism theorem says G/N {1, 1} Solution: We will see later in the course that any finite group consisting of p 2 elements, where p is a prime, is necessarily abelian. Here is a clever, yet somewhat brute force approach, for order 9. Since o(g) 9, Lagrange tells us that any element in G has order 1, 3, or 9. If any element has order 9 then G would be cyclic hence abelian, so we might as well assume that every non-identity element in G has order 3. Let L be any subgroup of G. The fact that every non-identity element of G has order 3 implies that for any g G we have g L if and only if g 2 L. If g L then trivially g 2 L. Conversely, if g 2 L then so is (g 2 ) 2 g 4 g 3 g g. Now let x be any non-identity element in G, then H (x) {e, x, x 2 }. Lagrange tells us that H has three cosets in G. I claim that if y G H, then the three cosets of H are H, Hy and Hy 2. By our above observation, neither y nor y 2 are in H so Hy H and Hy 2 H. Furthermore, the fact that y H implies as well that Hy Hy 2. Thus our 9 elements in G are of the form H {e, x, x 2 } Hy {y, xy, x 2 y} Hy 2 {y 2, xy 2, x 2 y 2 }. To show that G is abelian, it suffices to prove that xy yx since this would imply that any two elements of G commute. To attack this, let s first prove that yx Hy. Note that yx H since y H. If Hy 2 Hyx then necessarily yx(y 2 ) 1 yxy H. But then yxyx (yx) 2 H as well. By our above observation we then get that yx H, which is absurd. Thus yx Hy, so we have narrowed down our case work. Clearly yx y since x e. If yx x 2 y then we can multiply both sides by x on the left to get xyx y. But then this equation implies (yx) 2 yxyx y 2 xyxy (xy) 2 and we can square both sides to get xy yx. This cannot happen since then xy x 2 y implies that x e. Thus the only possibility we are left with is yx xy, which in turn shows that G is abelian. Page 3

4 Solution: By Lagrange s theorem, any element in G can only have order 1,2,3 or 6. G cannot have any elements of order 6, for otherwise G would be cyclic hence abelian. Thus every non-identity element of G has order 2 or 3. You proved in Herstein that if every element of a group is equal to its inverse (meaning it has order 1 or 2) then the group is abelian. This forces G to have an element y G such that o(y) 3. Let H (y) {e, y, y 2 } and choose x G H. Lagrange s theorem tells us that H has index 2 in G so the cosets of H are H {e, y, y 2 } Hx {x, yx, y 2 x}. As H has index 2 in G it must be normal so xh Hx. As G is non-abelian we cannot have x and y commuting (for otherwise G would be abelian since x and y generate G) so the fact that xh Hx yields yx xy 2. To show that G is isomorphic to S 3, the last thing we need to verify is that o(x) 2. Of course G/H is a group of order 2 so He (Hx) 2 Hx 2 which means that x 2 H. Now o(x) 2 or 3, but if o(x) 3 then x 3 e H would imply that x x 3 x 2 H which is absurd. Thus o(x) 2 and we see that G is isomorphic to S Solution: Let G be abelian. Then any subgroup N G is normal so G/N is a group. For any a, b G we know ab ba so in fact implies that G/N is abelian. NaNb Nab Nba NbNa Solution: Let G C {0} and let G be the set of matrices of the form a b b a where a, b R are not both zero. Before we construct an explicit isomorphism between G and G, let s remind ourselves the multiplication rules. In G we have and in G a b b a d c (a + bi)(c + di) (ac bd) + (ad + bc)i ac bd ad + bc (ad + bc) ac bd These multiplication rules strongly hint at what our map should be. Define ϕ: G G by a b ϕ(a + bi). b a The multiplication rules we just wrote show that ϕ is a homomorphism. The map ϕ is clearly onto and is also one-to-one since ker(ϕ) {1}. Thus ϕ is the desired isomorphism. Page 4

5 Solution: Let ψ : S 1 S 2 be a one-to-one map. We d like to produce a one-to-one homomorphism ϕ: A(S 1 ) A(S 2 ). To do so, start by letting ψ(s 1 ) T S 2. Then ψ : S 1 T is a one-to-one correspondence. This is because ψ is known to be one-to-one and is certainly onto its image. Thus we can define the inverse function ψ 1 : T S 1. Now let f A(S 1 ). We need to define ϕ(f) to be a one-to-one correspondence from S 2 to itself. I propose we define the function ϕ(f) as follows { z, z T ϕ(f)(z) ψ f ψ 1 (z), z T. Notice that this truly defines a function from S 2 to itself, so it only remains to check that ϕ(f) is actually a correspondence and that ϕ is a homomorphism and that ϕ is one-to-one. To prove that ϕ(f) is a one-to-one correspondence, it suffices to exhibit an inverse function. We define g : S 2 S 2 by { z, z T g(z) ψ f 1 ψ 1, z T. Notice that g is well-defined since f is invertible on S 1 and it is a simple matter to check that ϕ(f) g g ϕ(f) I, the identity map on S 2. This proves that ϕ(f) is actually an element of A(S 2 ). Next we need to show that ϕ is a homomorphism. Let f 1, f 2 A(S 1 ). For any z S 2 T we have by definition ϕ(f 1 f 2 )(z) z ϕ(f 1 ) ϕ(f 2 )(z). For z T we get ϕ(f 1 f 2 )(z) ψ f 1 f 2 ψ 1 (z) (ψ f 1 ψ 1 ) (ψ f 2 ψ 1 )(z) ϕ(f 1 ) ϕ(f 2 )(z). This proves that ϕ(f 1 f 2 ) ϕ(f 1 ) ϕ(f 2 ) and so ϕ is a homomorphism. The last thing to check is that ϕ is one-to-one, but this can be done by computing the kernel of ϕ. A function f is contained in ker(ϕ) iff ϕ(f)(z) z for every z S 2. This only occurs when ψ f ψ 1 (z) z for every z T, but this forces f to be the identity map on S 1. Thus the kernel of ϕ consists of the identity element of A(S 1 ) alone and hence ϕ is one-to-one. 2 For Fun Solution: Let G be the group of real numbers under addition, and let G denote the group of Page 5

6 complex numbers having absolute value 1 under multiplication. Define a function ϕ: G G by Note that ϕ(x) G since ϕ(x) e 2πix. e 2πix cos(2πx) + i sin(2πx) cos 2 (2πx) + sin 2 (2πx) 1. It is easy to check that ϕ is a homomorphism since ϕ(x + y) e 2πi(x+y) e 2πix+2πiy e 2πix e 2πiy ϕ(x)ϕ(y). Now x ker(ϕ) if and only if ϕ(x) e 2πix 1. But 1 e 2πix cos(2πx) + i sin(2πix) implies that cos(2πix) 1 and sin(2πix) 0. This occurs if and only if x Z. This proves that ker(ϕ) N. The last thing to prove is that ϕ is onto. This is obvious if you know properties of complex numbers. If not, we prove it as follows. Let a + bi G, then a 2 + b 2 1. Then surely 1 a 1 so there exists r R so that a cos(r). But then b 2 1 a 2 1 cos 2 (r) sin 2 (r). This proves that b ± sin(r). Notice, however, that cos(r) cos( r) while sin(r) sin( r). Thus we can change r to r if necessary to ensure that a cos(r) and b sin(r). Letting x r 2π shows that ϕ(x) e 2πix e ir cos(r) + i sin(r) a + bi. Thus ϕ is onto, and the result follows by the isomorphism theorems Solution: In problem 18 it is easy to prove that G N. Let { } a b G a, b, c, d R, ad bd 0 We define a function det: G R {0} by a b det ad bc. Then we have previously proven that det is a homomorphism. Since the image of det is abelian, problem 5 tells us that G ker(det). Of course ker(det) is just N from problem 18. We are tasked with proving that in fact N G. The idea is to find a subset S N for which S N G and that N (S). This will then prove that N G. This might be tricky, but if you are familiar with row reduction you might think to try the following. Let S consist of all matrices of the form where a 0, and all matrices of the form 1 x a 0 /a and x 1 Page 6

7 where x R. It is then clear that S N. Let a b N. If a 0 then you can verify that a b ( a 0 c/a 1 /a ) ( 1 b/a If a 0, then certainly c 0, and you can check 0 b 1 1 c (b + d)/c. 1 1 /c This computation shows that N (S) since we just gave a recipe for writing elements of N as products of elements from S. Thus it only remains to show that S G. You should verify that 1 x and x 1 ( 2 0 ( x 0 and a 0 /a ). ) 1 ( 1 x 2 x ) 1 ) 1 ( 1/2 /2 0 x 0 ( This proves that S G so in fact N (S) G. ) 1 ) 1 1. a 0 a 0 It s worth pointing out that the diagonal matrices are unnecessary. In fact, every 2 2 matrix of determinant 1 is a product of matrices of the form 1 x and. x 1 Page 7