Finitary symmetric group is normal in symmetric group

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This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Finitary symmetric group (?)) satisfying a particular subgroup property (namely, Normal subgroup (?)) in a particular group or type of group (namely, Symmetric group (?)).


Let S be a set. Let K = \operatorname{Sym}(S) be the symmetric group on S, and G be the finitary symmetric group on S. In other words, G is the subgroup of K comprising the finitary permutations, i.e., the permutations that move only finitely many elements. Then, G is a normal subgroup of K.

Related facts


Given: A set S. G = \operatorname{FSym}(S) is a subgroup of K = \operatorname{Sym}(S). \sigma \in K and \alpha \in G.

To prove: \sigma\alpha\sigma^{-1} \in G.

Proof: If \sigma \in K and \alpha \in G, we have:

(\sigma\alpha\sigma^{-1})(\sigma(x)) = \sigma(\alpha(x)).

Thus, \sigma(x) is moved by \sigma \alpha \sigma^{-1} if and x is moved by \alpha. Since \sigma is a permutation, this shows that the number of points moved by \alpha and \sigma\alpha\sigma^{-1} is equal. In particular, \sigma\alpha\sigma^{-1} also moves only finitely many points, and hence is in G.

Thus, G is normal in K.