Difference between revisions of "Finitary symmetric group is normal in symmetric group"

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(Created page with '{{subgroup property satisfaction| group = symmetric group| subgroup = finitary symmetric group| property = normal subgroup}} ==Statement== Let <math>S</math> be a set. Let <mat...')
 
 
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Let <math>S</math> be a set. Let <math>K = \operatorname{Sym}(S)</math> be the [[symmetric group]] on <math>S</math>, and <math>G</math> be the [[finitary symmetric group]] on <math>S</math>. In other words, <math>G</math> is the subgroup of <math>K</math> comprising the finitary permutations, i.e., the permutations that move only finitely many elements. Then, <math>G</math> is a [[normal subgroup]] of <math>K</math>.
 
Let <math>S</math> be a set. Let <math>K = \operatorname{Sym}(S)</math> be the [[symmetric group]] on <math>S</math>, and <math>G</math> be the [[finitary symmetric group]] on <math>S</math>. In other words, <math>G</math> is the subgroup of <math>K</math> comprising the finitary permutations, i.e., the permutations that move only finitely many elements. Then, <math>G</math> is a [[normal subgroup]] of <math>K</math>.
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==Related facts==
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* [[Finitary symmetric group is characteristic in symmetric group]]
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* [[Finitary symmetric group is automorphism-faithful in symmetric group]]
  
 
==Proof==
 
==Proof==

Latest revision as of 18:41, 5 April 2009

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 (?)).

Statement

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

Proof

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.