Oops, all I meant was that each of Im(L) and Im(R) lies *in* the centralizer of the other, since R and L commute with one another. I don't know why each should be the entire centralizer of the other. --Dan On 2013-12-26, at 11:38 AM, Dan Asimov wrote:
I think the centralizer statement is true just because right and left multilplication 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 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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