Math 210A: Algebra, Homework 5


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1 Math 210A: Algebra, Homework 5 Ian Coley November 5, 2013 Problem 1. Prove that two elements σ and τ in S n are conjugate if and only if type σ = type τ. Suppose first that σ and τ are cycles. Suppose further that σ and τ are conjugate so that σ = ρτρ 1. By earlier considerations, if τ = (a 1... a m ), where a i {1,..., n}, then σ = ρτρ 1 = (ρ(a 1 )... ρ(a m )). Hence σ and τ have the same number of symbols in their cycle, so type σ = type τ. Conversely, if type σ = type τ, then write σ = (a 1... a m ) and τ = (b 1... b m ). Define ρ S n so that ρ(a i ) = b i for each i {1,..., m}. Such a ρ is guaranteed to exist since we do not have ρ(a i ) = ρ(a j ) for any i j as b i b j for any i j. BY the above, ρσρ 1 = (ρ(a 1 )... ρ(a m )) = τ, so these cycles are conjugate. Now suppose that σ and τ are two permutations. Then we may write σ = σ 1 σ l, τ = τ 1 τ m where σ i, τ j are disjoint cycles. Then if σ = ρτρ 1, we have σi = ρ ( τj ) ρ 1 = ρτ j ρ 1. By the above reasoning, we must have type σ i = type τ j for some reordering of the cycles, so that σ and τ have the same type. Conversely, if type σ = type τ, then we have l = m and type σ i = type τ i for an appropriate rearrangement. Therefore For each there exists ρ i such that σ i = ρ i τ i ρ 1 i. Since the τ i are disjoint, the ρ i may be chosen disjoint so that ρ i ρ j = ρ j ρ i and ρ j τ i ρ 1 j = τ i for i j. Let ρ = ρ i. Then ρτρ 1 = ρτ j ρ 1 = ρ j τ j ρ 1 j = σ j = σ. Therefore the cycles are conjugate, so we are done. Problem 2. Prove that S 4 acts by conjugation on the set of nontrivial elements of the normal subgroup N = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. Using this action prove that S 4 /N = S 3. 1
2 It is clear that N is a subgroup of S 4 and, since it is comprised of full conjugacy classes, it is normal. Therefore the action of S 4 on N is welldefined. Further, since conjugation preserves cycle type (see problem 1), the conjugate of any 22cycle is also a 22cycle, so S 4 acts in particular on the nontrivial elements of N, which are the only 22cycles. Therefore there is a homomorphism ϕ from S 4 to S(X) = = S 3, where X = N \ {e}. First, we calculate the kernel of ϕ. We claim that it is N. Since N is an abelian subgroup, we have στσ = τ for every στ N. Therefore we have N ker ϕ. Further, ϕ is not trivial since S 4 itself is not abelian, so ker ϕ S 4. By previous considerations A 4 and N are the only proper normal subgroups of S 4. Suppose that ker ϕ = A 4. Then conjugation by (1 2 3) would be trivial for every element of N. However, (1 2 3)(1 2)(3 4)(3 2 1) = (1 4)(2 3). Therefore ker ϕ A 4, so we must have ker ϕ = N. Then since S 4 /N = 24/4 = 6 = S 3, we must have im ϕ = S 3. By the first isomorphism theorem, S 4 /N = S 3, and we are done. Problem 3. Let σ = (1 2 n) S n. Show that the conjugacy class of σ has (n 1)! elements. Show that the centraliser of σ is the cyclic subgroup generated by σ. As in problem 1, all ncycles are conjugate in S n. An ncycle is uniquely determined by ordering the symbols 1,..., n up to rearrangement. There are n! orderings of the symbols and n rearrangements for each ncycle, so in total there are (n 1)! ncycles. Therefore the conjugacy class of σ has (n 1)! elements. We now use the orbitstabiliser theorem. Let G = S n. Let Gσ be the orbit of σ and G σ its stabiliser. Then Gσ = G : G σ = (n 1)! = n!/ G σ = G σ = n. Since the stabiliser under conjugation is the centraliser, we know the centraliser of σ has n elements. Since σ commutes with itself, we have σ G σ. But since σ = n, we must have σ = G σ, so we are done. Problem 4. Let H be a proper subgroup of a finite group G. Suppose that G does not divide G : H!. Then G contains a nontrivial normal subgroup N such that N is a subgroup of H. In particular, G is not simple. Let G act on the cosets G/H by left translation. This induces a homomorphism ϕ : G S(G/H). If ϕ were injective, then im ϕ = G would be a subgroup of S(G/H). By Cayley s theorem, we would have G G : H!, which is false by assumption. Hence ϕ is not an injection, so ker ϕ is a nontrivial normal subgroup of G. Further, if g ker ϕ, we have gh = H, hence g H. Therefore ker ϕ is a normal subgroup of G contained in H. This completes the proof. 2
3 Problem 5. Prove that all groups of order 2p n and 4p n are not simple. If p is an odd prime, then there exists a Sylow psubgroup P of order p n. Then if G = 2p n, then G : P = 2, so P is normal, hence the group is not simple. Now let G = 4p n and suppose P is not normal in G. Then there exists another Sylow psubgroup Q such that P Q. Let H = P Q. Then we have 4p n P Q = P Q H = p2n H. Therefore H p n /4 > p n 2. Further, H is a pgroup and H < p n so we must have H = p n 1. Therefore since P : H = Q : H = p is the smallest prime dividing p n, we must have H P and H Q. Hence P Q N G (H), the normaliser of H. But as above we have calculated P Q = p2n p n 1 = pn+1 > 2p n = P Q = 4p n. Therefore N G (H) = G, so H G. Therefore G is not simple. If p = 2, then groups of order 2p n and 4p n are 2groups. By previous considerations, pgroups have nontrivial centres. Therefore groups of order 2p n and 4p n are not simple. This completes the proof. Problem 6. (a) Let N G be a subgroup. Prove that if H is contained in the centre of G and G/N is cyclic, then G is abelian. (b) Prove that any group of order p 2 is abelian. We follow the solution for Problem 5 on Homework 2. (a) We use the definition N G if and only if xnx 1 N for all x G. Let g N. Then xgx 1 = xx 1 g = g, since g Z(G), and hence xgx 1 N for all x G, g N. Hence N is normal. We claim that if G/N is cyclic that G is abelian. Write G/N = xn. Now let a, b G such that a (xz(g)) m = x m Z(G) and b x n Z(G). Then a = x m g 1 and b = x n g 2 for g 1, g 2 N. Then ab = (x m g 1 )(x n g 2 ) = x m (g 1 x n )g 2 = x m (x n g 1 )g 2 = (x m x n )(g 1 g 2 ) = (x n x m )(g 2 g 1 ) = x n (x m g 2 )(g 1 ) = x n (g 2 x m )g 1 = (x n g 2 )(x m g 1 ) = ba. We use the fact that g 1, g 2 Z(G) and that x commutes with itself freely. This shows that G is abelian. 3
4 (b) Let G act on itself by conjugation. Then by the class equation p 2 = G = Z(G) + G : N G (g). nontrivial conjugacy classes By Lagrange s theorem, G : N G (g) = G / N G (g) and since these conjugacy classes are proper and nontrivial, p G : N G (g). Since p p 2, we must have p Z(G), so Z(G) is nontrivial. If Z(G) = p 2, we are done. Suppose Z(G) = p. By 5(a), Z(G) is normal so we may examine the factor group G/Z(G). Since G/Z(G) = G / Z(G) = p, by earlier considerations G/Z(G) must be the cyclic group of order p, which implies that G is abelian, contradicting Z(G) = p. Therefore Z(G) = G and we are done. Problem 7. Let G be a nonabelian group of order p 3. Prove that the centre of G coincides with [G, G]. By earlier considerations, pgroups have nontrivial centres. Therefore Z(G) has order p or p 2 (since it is nonabelian). Suppose that Z(G) has order p 2. Then G/Z(G) = Z/pZ is cyclic, so by an earlier problem, G is abelian, which is a contradiction. Therefore we must have Z(G) = p. As such, G/Z(G) = p 2, and by the preceding problem, G/Z(G) is abelian. Since [G, G] has the property that it is the smallest subgroup N such that G/N is abelian, we must have [G, G] Z(G). But since Z(G) = p and [G, G] p (since G is nonabelian), we must have [G, G] = Z(G). This completes the proof. Problem 8. Let G be a semidirect product of a cyclic normal subgroup N of order n and an abelian group K. Show that if K is relatively prime to ϕ(n), then G is abelian. The semidirect product G = N K is given by a homomorphism ϕ : K Aut N such that nk n k = n ϕ k (n )kk where ϕ(k) = ϕ k Aut N. By earlier considerations, we know that Aut N = Aut Z/nZ = (Z/nZ). Further, (Z/nZ) = ϕ(n). By Cayley s theorem, im ϕ divides ϕ(n), and further im ϕ divides K. Therefore im ϕ divides gcd( K, ϕ(n)) = 1, so ϕ must be the trivial homomorphism. Therefore we have nk n k = nn kk for every n N, k K. Therefore N and K commute with each other, and since N and K are abelian, the whole group is abelian. This completes the proof. Problem 9. Determine the centre of D 2n. 4
5 First, the case of n = 1 is trivial since D 2 = Z/2Z. in the case that n = 2, we have D 4 = Z/2Z Z/2Z, so the group itself is abelian. Now let D 2n = r, s : r n = s 2 = 1, rs = sr 1. Let σ Z = Z(D 2n ). Suppose that σ = r m. Then r m sr k = r m k s? = r (m+k) s = sr m+k = sr k r m. This is only true if (k + m) m k (mod n) for all k, i.e. m m (mod n), which would mean m = n/2 (which only exists for even n). It is also clear that r m r k = r k r m for any k. Thus r m commutes with every element if m = n/2. Now suppose σ = s. Then s r r s in the case that n > 2. Hence s / Z. Finally, suppose that σ = sr m. Then sr m sr k = r m k? = r k m = sr k sr m. This is only true if m k k m (mod n) for all k, i.e. 2k = 2m (mod n) for all k, which is impossible if n > 2. Therefore no element of this form exists. Therefore we summarise the result: D 2n n = 1, 2 Z(D 2n ) = {e} n > 2, even {e, r n/2 } n > 2, odd Problem 10. Show that the exact sequence is not split. 0 Z Q Q/Z 0 Let i : Z Q and π : Q Q/Z be the injective and surjective maps in the exact sequence. We claim that Q = Q/Z Z as groups, so the sequence is not split. If the sequence is split, the isomorphism ϕ : Q Q/Z Z must be given by p/q (p /q, n) where 0 p /q < 1 and n + p /q = p/q so that i(n) = (0, n) and π(p/q) = p /q. Every element q Q has infinite order, so Q is torsionfree. However, every element of Q/Z is torsion, since an arbitrary element p/q has order q. In particular, let 1/n Q. Then (0, 1) = ϕ(1) = ϕ(n (1/n)) = n ϕ(1/n) = n (1/n, 0) = (0, 0), but this is absurd. Therefore such a ϕ cannot exist, so the sequence is not split. 5
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