[math-fun] Group theory question
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
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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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
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
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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On Thu, Dec 26, 2013 at 2:27 PM, Eugene Salamin <gene_salamin@yahoo.com>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?
If you want a reference you can start here: http://en.wikipedia.org/wiki/Holomorph_%28mathematics%29#Hol.28G.29_as_a_per...
participants (4)
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Dan Asimov -
Eugene Salamin -
Fred W. Helenius -
W. Edwin Clark