Excellent proof. Thank you. -- Gene
________________________________ From: Fred W. Helenius <fredh@ix.netcom.com> To: math-fun@mailman.xmission.com Sent: Thursday, December 26, 2013 12:21 PM Subject: Re: [math-fun] Group theory question
On 12/26/2013 2:49 PM, Eugene Salamin wrote:
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 ?
No, it is easy to see that any element of the centralizer of L(G) is in R(G). Suppose s is in the centralizer of L(G); that means s L(a) = L(a) s for any a in G. Applying both of these permutations to the identity 1 of G, we get s(a) = a s(1); that is, s is R(s(1)).
-- Fred W. Helenius fredh@ix.netcom.com
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