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Algebra 2. the Symmetric Groups Sn. Roma, December 15, 2009 in This Algebra 2. The symmetric groups Sn. Roma, December 15, 2009 In this note we determine the automorphism groups of the symmetric groups Sn. For n = 2 ∼ this is very easy: we have S2 = Z2 and hence Aut(S2) is trivial. Therefore we suppose from now on that n > 2. The main result is Theorem 8. For the convenience of the reader we first recall some basic properties of the groups Sn and the subgroups An of even permutations. Lemma 1. Let n > 2. (a) The center of Sn is trivial; (b) For n > 3 the center of An is trivial. Proof. (a) Let σ 2 Z(Sn). If σ 6= id, then there exist two distinct a; b 2 f1; 2; : : : ; ng with σ(a) = b. Choose c 2 f1; 2; : : : ; ng with c 6= a and c 6= b. Then (b c)σ 6= σ(b c) because (b c)σ maps a to c, while σ(b c) maps a to b. This shows that σ = id and Z(Sn) must be trivial, as required. (b) Similarly, suppose that σ 2 Z(An) is non-trivial. Pick two distinct a; b 2 f1; 2; : : : ; ng with σ(a) = b and choose two elements c; d 2 f1; 2; : : : ; ng different from a and b. Then (b c d)σ 6= σ(b c d) because the two permutations map a to different elements. Lemma 2. Two elements of Sn are conjugate if and only if they have the same cycle type. Proof. For any σ 2 Sn and any d ≤ n we have σ(1 2 : : : d)σ−1 = (σ(1) σ(2) : : : σ(d)): This shows that any conjugate of a d-cycle is again a d-cycle. Since every permutation is a product of disjoint cycles, it follows that the cycle types of conjugate permutations are the same. In the other direction, let τ = (a1 : : : ar)(ar+1 : : : as) ::: (al : : : am) and 0 0 0 0 0 0 0 τ = (a1 : : : ar)(ar+1 : : : as) ::: (al : : : am) be two permutations having the same cycle type. 0 Define σ 2 Sn by σ(ai) = ai for i = 1; 2; : : : ; m. Then −1 −1 −1 −1 στσ = σ(a1 : : : ar)σ σ(ar+1 : : : as)σ : : : σ(al : : : am)σ ; 0 0 0 0 0 0 = (a1 : : : ar)(ar+1 : : : as) ::: (al : : : am); = τ 0: This shows that τ and τ 0 are conjugate, as required. Lemma 3. Let n > 2. (a) The group An is generated by 3-cycles. (b) Any normal subgroup of An that contains a 3-cycle, is equal to An itself. Proof. (a) The product (1 2)(2 3) is equal to the 3-cycle (1 2 3). The product of two disjoint 2-cycles (a b) and (c d) is equal to (a b)(b c)(b c)(c d) and is hence a product of two 3-cycles. Since any element of An is a product of an even number of transpositions, it is therefore a product of 3-cycles. 0 (b) Let N ⊂ An be a normal subgroup and suppose that (1 2 3) 2 N. Let σ 2 An be an 0 −1 0 arbitrary 3-cycle. Then σ = τ(1 2 3)τ for some τ 2 Sn. If τ 2 An, then σ 2 N and we 0 0 0 0−1 are done. If not, then τ = τ(1 2) is in An and σ = τ (1 3 2)τ is once again in N. 1 Lemma 4. The commutator subgroup of Sn is equal to An. For n ≥ 5 the commutator subgroup of An is equal to An itself. 0 Proof. Since Sn=An is commutative, the commutator subgroup Sn is contained in An. −1 −1 0 Conversely, we have (1 2)(1 3)(1 2) (1 3) = (1 2 3), showing that every 3-cycle is in Sn. 0 By Lemma 3 (a) the group An is generated by 3-cycles, so that Sn = An as required. The identity (1 2 3)(3 4 5)(1 2 3)−1(3 4 5)−1 = (1 4 3): shows that for n ≥ 5 every 3-cycle is a commutator of An. This implies the second statement. We remark that the group A3 is abelian, so that its commutator subgroup is trivial. The group A4 is not abelian. Its commutator subgroup is V4 = f(1); (1 2)(3 4); (1 3)(2 4); (1 4)(2 3)g: Indeed, V4 is normal and the quotient A4=V4 has order 3 and is hence abelian. It follows 0 −1 −1 that A4 ⊂ V4. Equality follows from the identity (1 2 3)(1 2 4)(1 2 3) (1 2 4) = (1 2)(3 4). Proposition 5. Let n ≥ 5. Then the group An is simple, i.e. does not contain any proper normal subgroups. The only proper normal subgroup of Sn is An. Proof. Let N ⊂ An be a non-trivial normal subgroup. We will show that N contains a 3-cycle. Then Lemma 3 (b) implies the required result. Step 1. Suppose that N contains a permutation σ which is a product of disjoint cycles at least one of which has length d ≥ 4. Then, up to renumbering, we have σ = (1 2 : : : d)τ where τ leaves f1; 2; : : : ; dg invariant. The permutation σ−1(1 2 3)σ(1 2 3)−1 is contained in N. One easily checks that it is equal to the 3-cycle (1 3 d). Step 2. This leaves us with the possibility that all permutations in N are products of disjoint cycles of length ≤ 3. Suppose that N contains a permutation σ admitting a 3-cycle. If it admits only one 3-cycle, then its square is a 3-cycle and we are done. If it contains at least two 3-cycles, we may assume that σ = (1 2 3)(4 5 6)τ where τ leaves f1; 2;:::; 6g invariant. Then σ−1(1 2 4)σ(1 2 4)−1 is contained in N. One easily checks that it is equal to (1 4 2 3 6) and we are done by Step 1. Step 3. This leaves us with the possibility that all permutations in N are products of disjoint transpositions. Let σ 2 N be a non-trivial element. Since σ is even, it is a product of at least two transpositions and we may assume that σ = (1 2)(3 4)τ, where τ leaves f1; 2; 3; 4g invariant. Then σ(1 2 3)σ(1 2 3)−1 = (1 3)(2 4) is in N. Since n ≥ 5 the permutation (1 3)(2 4)(1 3 5)(1 3)(2 4)(1 3 5)−1 is in N. It is equal to the 3-cycle (1 3 5) and we are done. To prove the second statement of the Proposition, let N be a proper normal subgroup of Sn. Then N \ An is a normal subgroup of An. So either N ⊂ An in which case N = f1g or N = An or we have N \ An = f1g. In the latter case #N ≤ 2 and hence N ⊂ Z(Sn). Lemma 1 implies then that N = f1g. This proves the proposition. We remark that the possibility that arises in Step 3 of the proof of Lemma 5, actually occurs for n = 4. In that case the group V4 mentioned above is a normal subgroup of A4. Its elements are products of disjoint transpositions. 2 Corollary 6. For n ≥ 5, the proper subgroups of An have index at least n. Proof. Let H ⊂ An be a subgroup of index m. Translation of the left cosets of H gives rise to a non-trivial homomorphism An −! Sm. By Lemma 4 the group An admits no 1 proper normal subgroups, so that the map is injective. This implies 2 n! ≤ m!, which can only happen when n ≤ m as required. Lemma 7. Let n > 2 and let F : Sn −! Sn be an automorphism mapping transpositions to transpositions. Then F is an inner automorphism. Proof. The product of two distinct transpositions has order 2 when the transpositions are disjoint, while it is a 3-cycle and thus has order 3 when they are not. Therefore any automorphism of Sn maps pairs of disjoint transpositions to pairs of disjoint transpositions. Let F (1 2) = (a b). Let x 2 f1; 2; : : : ; ng be different from 1 or 2. Since (1 2)(1 x) is a 3-cycle, so is F (1 2)F (1 x) = (a b)F (1 x). It follows that the 2-cycle F (1 x) moves either a or b. Switching a and b if necessary, we may assume that it moves a, so that F (1 x) = (a c) for some c different from a and b. Claim. For every y 6= 1 we have F (1 y) = (a d) for some d 2 f1; 2; : : : ; ng different from a. Proof of the claim. This is clear when y = 2 or y = x, so we may assume y 6= 2; x. Since both permutations (1 y)(1 2) and (1 y)(1 x) are 3-cycles, so are their images under F . We have F (1 2) = (a b) and F (1 x) = (a c). Therefore, if F (1 y) is not moving a, then it must move both b and c. Since F (1 y) is a transposition, this means that F (1 y) = (b c). Applying F −1 to the relation (a b)(a c)(b c) = (a c); we find (1 2)(1 x)(1 y) = (1 x): Since y 62 f1; 2; xg, the permutation on the left maps 1 to y.
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