Automorphisms of S n and of A n

Transcription

1 Automorphisms of S n and of A n In this note we prove that if n 6, then Aut(S n ) = S n = Aut(An ). In particular, when n 6, every automorphism of S n is inner and every automorphism of A n is the restriction of an inner automorphism of S n. However, Aut(S 6 ) is not isomorphic to S 6 ; while we do not prove it, in fact, Aut(S 6 ) = Aut(A 6 ) satisfies [Aut(S 6 ) : Inn(S 6 )] =. See Chapter. of [1] for this fact. Recall that S n and A n have trivial center (if n for S n and n for A n ). Our arguments will be combinatorial. For that reason, we mention some relevant properties of S n. First, the conjugacy class of a permutation σ is the set of all permutations with the same cycle structure as σ. Second, S n is generated by the transpositions {(1, i) : i > 1}. To see this we note that (1, r)(1, s)(1, r) = (r, s) if r s; this proves the claim since S n is generated by the transpositions. Lemma 1. If 1 k n/, then the number of products of k disjoint transpositions in S n is n!/ ( k k!(n k)! ). Proof. A product of k disjoint transpositions in S n has the form (a 1, b 1 ) (a k, b k ) with the a i, b i distinct integers between 1 and n. Choosing a single transposition (a, b) can be done in n(n 1)/ ways; we note that (a, b) = (b, a). We have n(n 1)/ choices for (a 1, b 1 ). Similarly, there are (n )(n )/ choices for (a, b ) once (a 1, b 1 ) has been chosen. Continuing this argument, and keeping track of order of the (a i, b i ), there are n(n 1) (n )(n ) (n k )(n k 1) = n! k (n k)! choices for an ordered list of k disjoint transpositions. Since order doesn t matter, we must divide by k! to get the possible products that result. Thus, the formula stated in the lemma is true. Lemma. Let ϕ Aut(S n ). If ϕ sends transpositions to transpositions, then ϕ is inner. Proof. Suppose that ϕ(1, r) = (a r, b r ) for each r. Then ϕ ((1, )(1, r)) = (a, b )(a r, b r ). However, if r, then (1, )(1, r) = (1, r, ), an element of order. Thus, either a r {a, b } or b r {a, b }. By reversing a r and b r if necessary, we may suppose that a r {a, b } for all r. We claim that either a r = a for all r or a r = b for all r; suppose instead that there are r s with a r = a and a s = b. Note that (1, r, )(1, s, ) = (1, s)(, r) has order. 1

2 However, ϕ ((1, r, )(1, s, )) = (a, b )(a r, b r )(a, b )(a s, b s ) = (a, b )(a, b r )(a, b )(b, b s ) = (a, b r, b )(a, b, b s ) = (b, b s, b r ) has order. This is a contradiction. Thus, we must either have a = a r for all r or b = b r for all r. We assume that a = a r for all r; the other case is similar. We then have ϕ(1, r) = (a, b r ) for all r. Note that this forces b r b s if r s since ϕ is 1-1. Let x be a permutation for which x(1) = a and x(r) = b r for all r. This uniquely determines x; we have defined x on n 1 values, which is enough to completely determine a permutation of n elements. From the choice of x we see that ϕ(1, r) = (a, b r ) = x(1, r)x 1. Therefore, ϕ = Int(x) is inner. Theorem. If n 6, then every automorphism of S n is inner. Thus, Aut(S n ) = S n. Proof. Let ϕ Aut(S n ). If σ is a transposition, then ϕ(σ) has order. Thus, ϕ(σ) is the product of k 1 disjoint transpositions for some k. Now, ϕ sends conjugacy classes to conjugacy classes. The conjugacy class of the product of k disjoint transpositions is the set of all products of k disjoint transpositions. Thus, Lemma 1 implies that n(n 1)/ = n!/ ( k k!(n k)! ). We note that this equation is valid if k = 1 or if n = 6 and k =. It is easy to check that it is not valid if n < 6. We rewrite the equation as (n )! = k 1 k!(n k)! We first show, by induction on k, that if n = k 8, then (n )! > k 1 k! This is clear for k = 4. Suppose that it is true for k. Then ((k + 1) )! = (k)! = k(k 1)(k )! > k(k 1) k 1 k! = k k(k 1)k! > k (k + 1)! since k(k 1) k + 1. Thus, this claim is true. Next we show, by induction on n, that if n 7, then (n )! > k 1 k!(n k)! whenever n > k. It is easy enough to verify this for n = 5. Suppose that the result is true for n. If n + 1 > k, then either n = k or n > k. If n = k, then k 4, and by the previous paragraph, we have (n 1)! = (n 1)(n )! > (n 1) k k! = (n 1) k k!(n k)! > k k!(n k)! On the other hand, if n > k, then the induction hypothesis yields (n 1)! = (n 1)(n )! > (n 1) k k!(n k)! > k k!(n + 1 k)!

3 This finishes the proof of this second claim. What we have proven is that, if n 6, then the conjugacy classes of transpositions and products of k disjoint transpositions are of different sizes. Thus, ϕ sends transpositions to transpositions. By Lemma, this implies that ϕ is inner. We now consider A n. Recall that A n is generated by -cycles. For an easy proof of this, we note that A n is generated by products of transpositions. Since (a, b)(c, d) = (a, c, b)(a, c, d), (a, b)(b, c) = (a, c, b) whenever a, b, c, d are distinct, the claim is verified. To help with the following proof, we note that there are four possibilities for the product of two -cycles: (a, b, c)(a, b, d) = (a, d)(b, c), (a, b, c)(a, d, b) = (b, c, d), (a, b, c)(a, d, e) = (a, b, c, d, e), (a, b, c)(d, e, f). The only case where we get an element of order is in the first case. Lemma 4. The number of products of k disjoint -cycles is n!/ ( k k!(n k)! ). Proof. The argument is similar to that of Lemma. If σ = (a 1, b 1, c 1 ) (a k, b k, c k ), then the number of choices for a 1, b 1, c 1 is n(n 1)(n ), and the same cycle is represented in three ways. Repeating this idea, we see that the number of choices for an ordered list τ 1 τ k of disjoint -cycles is n(n 1)(n ) (n k + )(n k + )(n k + 1) = n! k (n k)! Since order of the product τ 1 τ k does not change the permutation, we must divide by k! to count the number of these products. This proves the lemma. Lemma 5. Let ϕ Aut(A n ). If ϕ sends -cycles to -cycles, then ϕ = Int(x) An x S n. for some Proof. Let u i = (1,, i). We claim that there are a 1, a so that for each i, we have ϕ(u i ) = (a 1, a, a i ) for some a i. Set v i = ϕ(u i ). Note that u i u j has order whenever i j by the calculation before Lemma 4. Thus, v i v j also has order. Therefore, there are a 1, a with v = (a 1, a, c) and v 4 = (a 1, a, d). Consider v i for i 5. If v i fixes a 1, then we must have v i = (a, c, ) and v i = (a, d, ). This is impossible. Therefore, a 1 occurs in the cycle structure of v i, and this forces v i = (a 1, a, a i ). This proves our claim. Define x S n by x(i) = a i for all. Then xu i x 1 = v i by a short calculation. Thus, ϕ = Int(x) An, as desired.

4 To prove the following theorem, we relate conjugacy classes in S n to those in A n. If σ A n, then its conjugacy class Cl Sn (σ) has order [S n : C Sn (σ)], where C Sn (σ) is the centralizer of σ in S n. Similarly, Cl An (σ) has order [S n : C An (σ)]. Now, C An (σ) = A n C Sn (σ). From this and the counting formula N H NH = N H if N and H are subgroups of a groups G with N normal, we conclude that Cl An (σ) = Cl Sn (σ) or Cl An (σ) = 1 Cl S n (σ). The first case occurs when C An (σ) A n, and the second case occurs otherwise. If σ = (a, b, c) is a -cycle and n 5, then (1, ) commutes with σ and lies outside of A n. Thus, Cl An (σ) = Cl Sn (σ). If σ = τ 1 τ k is a product of k disjoint -cycles, write τ 1 = (a, b, c) and τ = (d, e, f). Then (a, d)(b, e)(c, f) commutes with σ; thus, we have Cl An (σ) = Cl Sn (σ) also in this case. Theorem 6. If n and n 6, then every automorphism of A n is the restriction of an inner automorphism of S n. Consequently, Aut(A n ) = S n. Proof. Let ϕ Aut(A n ). if σ is a -cycle, then ϕ(σ) has order ; thus, it is the product of k 1 disjoint -cycles. If n < 6, then k = 1 is the only possibility; the result then follows from Lemma 5. Thus, suppose that n 6. Since ϕ sends conjugacy classes to conjugacy classes, the conjugacy class in A n of a -cycle then has the same size as the conjugacy class of a product of k disjoint -cycles. By Lemma 4 and the comments immediately before the statement of the theorem, we then have n(n 1)(n )/ = n!/ ( k k!(n k)! ). This can be proven to occur only when n = 6 and k = by methods similar to the proof of Theorem. Therefore, Lemma 5 shows that ϕ is the restriction of an automorphism of S n, and the rest then follows from Theorem. For completness, we prove that (n )! > k 1 k!(n k)! whenever n > 6; this was the claim used in the proof of Theorem 6. We do this in two cases. First, if k, we prove that (k )! > k 1 k! by induction on k. The case k = is clear. Suppose the result is true for k. Then ((k + 1) )! = (k)! = (k)(k 1)(k )(k )! > (k)(k 1)(k ) k 1 k! = k k(k 1)(k )k! > k (k + 1)! Thus, by induction, this claim is true for all k. Next, we prove by induction on n, that if n 7 and n k, then (n )! > k 1 k!(n k)!. The result is easily seen to be true for n = 7. Thus, we assume that n 7 and that the result holds for n. If n > k, then by the induction hypothesis, (n )! = (n )(n )! > (n ) k 1 k!(n k)! k 1 k!(n + 1 k) 4

5 since k 1. On the other hand, if n = k, then (n )! = (n )(n )! > (n ) k 1 k! = (n ) k 1 k!(n + 1 k)! > k 1 k!(n + 1 k)! by the earlier claim. Thus, the result is true for all n 7 and all k with n k. References [1] Michio Suzuki, Group theory. I, Springer-Verlag, Berlin,