3.36:

Let A_infty be the infinite alternating group, by definition the subgroup generated by 3-cycles in the group S_infty of all permutations of the set of positive integers. (Note, any product of 3-cycles fixes all but finitely many points.) We are to show Let A_infty is simple. Suppose N is a nontrivial normal subgroup of A_infty . Let x be a nontrivial element of N. Then there is a finite set S, containing at least five elements, such that x fixes every point in the complement of S. Then x(S) = S. Let A_Sdenote the group of even permutations of S, considered as a subgroup of A_infty. (We know every even permutation of S is a product of 3-cycles, by a theorem from class.) Then x is in N ^ A_S (here ^ means intersection). Then N ^ A_S is nontrivial, and is normal in A_S , hence is equal to A_S, since S was assumed to have at least five elements, and A_S is isomorphic to A_|S|. Then N contains a 3-cycle (abc), where a, b, c are in S. If (def) is any other 3-cycle, define an element y of S_infty by y = (ad)(be)(cf), where (rs) is the identity if r = s. Choose two distinct numbers g and h different from a, b, and c, and let z = (gh). Then (abc) conjugated by y or yz is equal to (def), and one of y or yz is even (since they differ by a transposition), hence (abc) and (def) are conjugate in A_infty. Then, since (abc) is in N, and N is normal in A_infty , N contains every 3-cycle. Then N = A_infty . Thus A_infty is simple.