User talk:JackSchmidt/Archives/2009/08

Prosolvable Groups
Dear Jack,

thanks for going through my article on Prosolvable Groups. Of course you are absolutely right; the definition I had given first was complete bullshit, I have corrected it by now.

Thanks a lot, .:ENE:.--ElNuevoEinstein (talk) 18:53, 11 August 2009 (UTC)

Order of an element in a semidirect product of groups
Hi Jack. I have a little question for you and the other contributors of the semidirect product article. Suppose that we know the order of both n and h in their respective groups N and H; can we say something about the order of the element (n,h) in the semidirect product N times H?. Please excuse me for taking some of your time with this question, that maybe doesn't have much to do with the article, but I don't know who else to ask. Thank you very much in advance and best regards. —Preceding unsigned comment added by 110.32.178.54 (talk) 06:23, 16 August 2009 (UTC)
 * Assuming N is normal and H is the complement, for any positive integer k, (n,h)^k = (n',h^k), so the order of (n,h) is a multiple of the order of h. On the other hand, one can give a pretty explicit formula for n' = n*n^h*n^(h^2)*...*n^(h^(k-1)).  So n' is a product of conjugates of n.  If k is the order of h and N is abelian, then n' is actually centralized by h, so in a Frobenius group with abelian kernel, the order of (n,h) is exactly equal to the order of h unless h=1, in which case it is exactly equal to the order of n.  In general, the smallest upper bound you can get is fairly unimpressive: by Lagrange's theorem, the order of (n,h) divides the product of the order of N and the order of h, and this is actually obtained whenever N is cyclic of order coprime to the order of h and h acts trivially on N.  Probably you will want to use more specialized information about your particular semidirect product to get a better upper bound. JackSchmidt (talk) 14:31, 16 August 2009 (UTC)
 * As an example, consider the semidirect product where H=N is the symmetric group on three points, H acting by conjugation on N. Take n=(1,2) and h=(2,3).  Then (n,h)*(n,h) = (n*n^h,h^2) = ( (1,2,3),  ) has order 3.  So we took two elements of order 2 and put them together to get an element of order 6.  This shows that in my previous comment we cannot replace "N" by "the subgroup generated by n", though of course we can replace N by any subgroup of N normalized by h and containing n.  In this case though, the h-normal closure of n is all of N. JackSchmidt (talk) 16:17, 16 August 2009 (UTC)