ALGEBRA HW 4 CLAY SHONKWILER.2.19 Prove that if N is a nmal subgroup of the finite group G ( N, G : N ) = 1 then N is the unique subgroup of G of der N. Proof. Let H G such that H = N. By Proposition 1, NH = N H N H = N 2 N H = N N N H Since N is nmal, NH G by Collary 15, so NH divides G, On the other h, so Hence, N G : N G : N = m N N G = m NH = m N N H. N H. G = G : N N, = N m N N H cancelling yields N N H divides both N G : N so, by hypothesis, N N H = 1 Therefe, N H = N, which means, since H = N, that H = N. We conclude, then, that N is the unique subgroup of G with der N..4.2 Find all composition series f Q 8 all 7 composition series f D 8. List the composition facts in each case. Answer: 1 {1, 1} {1, 1, i, i} Q 8 1 {1, 1} {1, 1, j, j} Q 8 1 {1, 1} {1, 1, k, k} Q 8 where, in each case, N i+1 /N i = Z/2Z. 1 s s, r 2 D 8 1 r 2 s, r 2 D 8 1
2 CLAY SHONKWILER 1 r 2 r D 8 1 sr sr, sr D 8 1 sr sr, sr D 8 1 sr 2 s, sr 2 D 8 1 s s, sr 2 D 8 where, in each case, N i+1 /N i = Z/2Z..4.5 Prove that subgroups quotient groups of a solvable group are solvable. Proof. Let G be a solvable group with composition series 1 = G 0 G 1... G n = G. Let H G. Consider H G i H G i+1. Each is clearly a group. If g H G i h H G i+1, then hgh 1 H since g, h H. Also, hgh 1 G i, since G i G i+1, so we can conclude that H G i H G i+1. This also implies that H G i+1 N G (H G i ). Hence, we see that H is solvable, as we can construct the series 1 H G 1... H G n 1 H G n = H G = H..5.4 Show that S n = (12), (12... n) f all n 2. Proof. We want to show that f any transposition (pq) where p < q, (pq) (12), (12... n). Now,, in general, (12... n)(12)(12... n) 1 = (12... n)(12)(1n... 2) = (2) (12... n)(m(m + 1))(12... n) 1 = ((m + 1)(m + 2)). Furtherme, f a transposition (pq), (pq) = ((q 1)q)... ((p + 1)(p + 2))(p(p + 1))((p + 1)(p + 2))... ((q 1)q). Since each term on the right is generated by (12) (12... n), then so is (pq). Since our choice of transposition was arbitrary, we see that every transposition in S n is in (12), (12... n). Since, as we ve seen, every element of S n can be written as a product of transpositions, we see that, f all σ S n, σ (12), (12... n). Hence, S n = (12), (12... n).
ALGEBRA HW 4.5.10 Find a composition series f A 4. Deduce that A 4 is solvable. Answer: 1 (12)(4) (12)(4), (1)(24) A 4 is a composition series of A 4. To see that A 4 is, in fact, solvable, it suffices to note that (12)(4) /1 Z/2Z (12)(4), (1)(24) / (12)(4) Z/2Z A 4 / (12)(4), (1)(24) Z/Z, each of which is a simple abelian group. 1 Let G be a group. The opposite group, G op, is the group which is equal to G as a set, whose group law µ is defined by Prove that G op is isomphic to G. Proof. Define φ : G G op by µ (x, y) = µ G (y, x) x, y G op. φ(x) = x 1. This is well-defined since x 1 G is equal to x 1 G op. Then ker(φ){x G x 1 = 1} = {1} so φ is injective. Also, f any x G op, φ(x 1 ) = (x 1 ) 1 = x, so φ is surjective. Finally, f x, y G, φ(µ(x, y)) = µ(x, y) 1 = µ(y 1, x 1 ) = µ (x 1, y 1 ) = µ (φ(x), φ(y)), so φ is a homomphism. isomphism. Since φ is a bijective homomphism, it is an
4 CLAY SHONKWILER 2 Two homomphisms f 1, f 2 from a group G 1 to a group G 2 are conjugate if there exists an element g G 2 such that f 1 (x) = gf 2 (x)g 1 f all x G 1. (a) Find all homomphisms from S to C. Let f : S C be a homomphism. Then ker(f) S. We know, from the last homewk, that {1}, (12), (1), (2), (12), S comprises the entire list of subgroups of S. If ker(f) = {1}, then f is a monomphism, meaning S f(s ). However, S is not abelian, whereas C is, so this is impossible. If ker(f) = S, then f is just the trivial homomphism. Now, we know that f induces an isomphism f : S /ker(f) f(s ). If ker(f) = 2, then S /ker(f) Z/Z, meaning f(s ) is a cyclic subgroup of C of der. The only such group is the group of rd roots of unity, {1, e 2π, e 4π }. Hence, there are precisely two possibilities f f given ker(f). F example, if ker(f) = (12), then S /ker(f) = S / (12) = { (12), (1) (12), (2) (12) }. Then f ( (12) ) = 1, f ((1) (12) ) = e 2π f ((1) (12) ) = e 4π f ((2) (12) ) is whatever remains. Hence, we deduce that f(1) = f((12)) = 1, f((1)) = f((12)) = e 2π, f((2)) = f((12)) = e 4π f(1) = f((12)) = 1, f((1)) = f((12)) = e 4π, f((2)) = f((12)) = e 2π Similarly, if ker(f) = (1), then f(1) = f((1)) = 1, f((12)) = f((12)) = e 2π f(1) = f((1)) = 1, f((12)) = f((12)) = e 4π if ker(f) = (2), then f(1) = f((2)) = 1, f((12)) = f((12)) = e 2π., f((2)) = f((12)) = e 4π, f((2)) = f((12)) = e 2π, f((1)) = f((12)) = e 4π f(1) = f((2)) = 1, f((12)) = f((12)) = e 4π, f((1)) = f((12)) = e 2π Finally, if ker(f) =, then ker(f) = (12), so S /ker(f) Z/2Z. That is to say that f(s ) is a cyclic group of der two in C. The only such possibility is the group {1, 1}. This means f( (12) ) = 1 f((12) (12) ) = 1. Specifically, f(1) = f((12)) = f((12)) = 1, f((12)) = f((1)) = f((2)) = 1..
ALGEBRA HW 4 5 Therefe, the above constitute all possible homomphisms from S to C. (b) Determine all homomphisms from S to S up to conjugation. Clearly the trivial homomphism f(x) = 1 f all x S is one such. Replacing e 2π with (12) e 4π with (12) makes it clear that the only such homomphisms having (12) as their kernel are f g such that: f(1) = f((12)) = 1, f((1)) = f((12)) = (12), f((2)) = f((12)) = (12) g(1) = g((12)) = 1, g((1)) = g((12)) = (12), g((2)) = g((12)) = (12). However, f(x) = (12)g(x)(12) f all x S, so these two homomphisms are conjugate. A similar argument gives a single homomphism (up to conjugation) f each kernel (1) (2). However, homomphisms with different kernels of degree two will not be conjugate, as can be sen simply by noting that there is no x S such that x(12)x 1 = 1. Now, if f is a homomphism with kernel (12), then, paralleling our arguments in the previous part, we see that either f(1) = f((12)) = f((12)) = 1, f((12)) = f((1)) = f((2)) = (12) f(1) = f((12)) = f((12)) = 1, f((12)) = f((1)) = f((2)) = (1) f(1) = f((12)) = f((12)) = 1, f((12)) = f((1)) = f((2)) = (2). If we call the first of these possibilities f 1, the second f 2 the third f, then it is readily apparent that f 1 (x) = (2)f 2 (x)(2), f 1 (x) = (1)f (x)(1) f 2 (x) = (12)f (x)(12) f all x S, so f 1, f 2 f are conjugate. Finally, if the kernel of a homomphism from S to S is trivial, then that homomphism is, in fact, an automphism. In the last homewk, we saw that any automphism f of S is of the fm f(x) = axa 1 f some a S f all x S. Hence, if f g are two automphisms of S, then f(x) = axa 1 g(x) = bxb 1 f some a, b S. However, g(x) = bxb 1 = b(a 1 a)x(a 1 a)b 1 = (ba 1 )(axa 1 )(ab 1 ) = (ba 1 )f(x)(ba 1 ) 1, so f g are conjugate. Therefe, we conclude that, up to conjugation, there is exactly one homomphism of S into itself f each of the 6 possible kernels. (c) Prove that any two injective homomphisms from S to GL 2 (R) are conjugate.
6 CLAY SHONKWILER Proof. Let f g be monomphisms from S to GL 2 (R). Then f(s ) g(s ) are isomphic to S. Consider the map g 1 f : S S. Note that g 1 is an isomphism. If x, y S such that (g 1 f)(x) = (g 1 f)(y), then g 1 (f(x)) = g 1 (f(y)) so f(x) = f(y), meaning x = y, so g 1 f is injective. Since S is finite, g 1 f is clearly surjective. Also, if x, y S, (g 1 f)(xy) = g 1 (f(xy)) = g 1 (f(x)f(y)) = g 1 (f(x))g 1 (f(y)) = (g 1 f)(x)(g 1 f)(y), so g 1 f is an automphism. Hence, as shown in last week s homewk, there exists τ S such that f all x S. Now, (g 1 f)(x) = τxτ 1 f(x) = ((g g 1 ) f)(x) = (g (g 1 f))(x) = g((g 1 f)(x)) = g(τxτ 1 ) = g(τ)g(x)g(τ 1 ), so we see that f g are conjugate. Since our choice of f g was arbitrary, we conclude that any two monomphisms from S to GL 2 (R) are conjugate. DRL EA, University of Pennsylvania E-mail address: shonkwil@math.upenn.edu