Automorphism group of finitary symmetric group equals symmetric group

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Statement

Let S be an infinite set. Let K = \operatorname{Sym}(S) be the Symmetric group (?) on S, i.e., the group of all permutations on S, and G = \operatorname{FSym}(S) be the subgroup of K that comprises the finitary permutations. In other words, G is the Finitary symmetric group (?) on S.

Then, G is normal in K. Further consider the map \varphi: K \to \operatorname{Aut}(G) sending an element a \in K to conjugation by a on G. This map is an isomorphism.

Related facts

Facts used

  1. Finitary symmetric group is normal in symmetric group
  2. Finitary symmetric group is centralizer-free in symmetric group
  3. Conjugacy class of transpositions is preserved by automorphisms
  4. Transposition-preserving automorphism of finitary symmetric group is induced by conjugation by a permutation

Proof

  1. By fact (1), the map \varphi: K \to \operatorname{Aut}(G) is well-defined, and by fact (2), \varphi is injective.
  2. By fact (3) every element of \operatorname{Aut}(G) sends transpositions to transpositions. Combining this with fact (4), we obtain that every element of \operatorname{Aut}(G) comes from an inner automorphism in K. Thus, \varphi is surjective.