Symmetric groups on infinite sets are complete: Difference between revisions

Latest revision as of 18:31, 5 April 2009

Let $S$ be an infinite set. The symmetric group on $S$ , denoted $S y m (S)$ , is a complete group: it is centerless and every automorphism of it is inner.

Given: $S$ is an infinite set, $K = S y m (S)$ , $σ$ is an automorphism of $K$ .

To prove: $σ$ is inner.

Proof: Let $G = F S y m (S)$ be the subgroup of $K$ comprising the finitary permutations.

By fact (1), $σ$ restricts to an automorphism, say $τ$ of $G$ .
By fact (2), the automorphism $τ$ of $G$ arises from some inner automorphism, say $σ^{'}$ , of $K$ .
Consider the ratio $σ^{'} σ^{- 1}$ . The restriction of this automorphism to $G$ is $τ τ^{- 1}$ which is the identity map. By fact (3), $σ^{'} σ^{- 1}$ is the identity map on $K$ , so $σ = σ^{'}$ . Thus, $σ$ is inner.

@@ Line 5: / Line 5: @@
 ==Facts used==
-# [[uses::Symmetric groups are centerless]]
+# [[uses::Finitary symmetric group is characteristic in symmetric group]]
-# [[uses::Conjugacy class of transpositions is preserved by automorphisms]]
+# [[uses::Automorphism group of finitary symmetric group equals symmetric group]]
-# [[uses::Transposition-preserving automorphism of finitary symmetric group is induced by conjugation by a permutation]]
 # [[uses::Finitary symmetric group is automorphism-faithful in symmetric group]]
+==Proof==
+'''Given''': <math>S</math> is an infinite set, <math>K = \operatorname{Sym}(S)</math>, <math>\sigma</math> is an automorphism of <math>K</math>.
+'''To prove''': <math>\sigma</math> is inner.
+'''Proof''': Let <math>G = \operatorname{FSym}(S)</math> be the subgroup of <math>K</math> comprising the finitary permutations.
+# By fact (1), <math>\sigma</math> restricts to an automorphism, say <math>\tau</math> of <math>G</math>.
+# By fact (2), the automorphism <math>\tau</math> of <math>G</math> arises from some inner automorphism, say <math>\sigma'</math>, of <math>K</math>.
+# Consider the ratio <math>\sigma'\sigma^{-1}</math>. The restriction of this automorphism to <math>G</math> is <math>\tau\tau^{-1}</math> which is the identity map. By fact (3), <math>\sigma'\sigma^{-1}</math> is the identity map on <math>K</math>, so <math>\sigma = \sigma'</math>. Thus, <math>\sigma</math> is inner.