ab=x°2 => ba = b(ab)b°-1 = (bxb°-1)(bxb°-1)=y°2 Cheers, Hartmut Dan Hoey <Hoey@aic.nrl.navy.mil>@mailman.xmission.com on 26.09.2003 11:39:12 Bitte antworten an math-fun <math-fun@mailman.xmission.com> Gesendet von: math-fun-bounces@mailman.xmission.com An: John Conway <conway@math.princeton.edu> Kopie: math-fun <math-fun@mailman.xmission.com> Thema: Re: [math-fun] non-square products of squares? This really is fascinating. I've been trying to follow along with Gap, and I ran into a problem because it takes about five or ten minutes to find all the elements of Aut(2^4). I kept trying to figure out what was wrong, where if I waited I would have had it. So now I can calculate semidirect products 2^4:3[h], where h is a homomorphism from 3 to Aut(2^4), and there are the three you promised: The canonical 2^4:3, the direct product 2^4x3, and... well, one that's halfway in between. I don't know if it's worth giving it a special name. Anyway, I still haven't found any group with a greater proporition of (a,b,ab)=(square,square,nonsquare) than 1/6, except for the 5/24 we get from 2^4:3. In fact, the proportion seems to go down, so perhaps 2^4:3 is unique. I have no idea how one might prove it, though. Here's a poser that came up while I was writing the code to search for nonsquare products. Can it ever happen that (a,b,ab,ba)=(square,square,square,nonsquare)? I can prove it's impossible for finite groups, but what about infinite? Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun