But is there a permutation of G that commutes with all the L(a), and is not one of the R(b) ? That is, can the centralizer of im L be bigger than im R ? -- Gene
________________________________ From: Dan Asimov <dasimov@earthlink.net> To: Eugene Salamin <gene_salamin@yahoo.com>; math-fun <math-fun@mailman.xmission.com> Sent: Thursday, December 26, 2013 11:38 AM Subject: Re: [math-fun] Group theory question
I think the centralizer statement is true just because right and left multiplication commute with one another in general.
--Dan
On 2013-12-26, at 11:27 AM, Eugene Salamin wrote:
Let G be a group, and let S be the permutation group on G. For a in G, let L(a) be the permutation that sends x to ax, and let R(b) be the permutation that sends x to xb' (' denoting inverse). Then im L and im R are isomorphic copies of G in S, and each is the centralizer in S of the other. I have a vague memory of having seen a proof of the statement concerning centralizers, but can't pin it down. Can someone point me to either a proof, or a counterexample?
-- Gene