26 Dec
2013
26 Dec
'13
1:21 p.m.
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