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.

It's 2^2 x A(4).

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.

Since the two best examples are A(4) = 2^2 : 3 and 2^4 : 3 it would seem that 2^6 : 3 deserves examination. Let me try to do that now (despite the fact that I have to lecture in a few minutes). The squares of the elements of order 3 ARE the elements of order 3, and the only other square is 1. So "(sq,sq,non-sq)" means "orders are (3,3,2)". Now any element of order 3 times an element of order 2 is of order 3, so the number of such triples is #3 x #2, where #n is the number of elements of order n. This makes it 128 x 63, as compared to 192 x 192, for a ratio of 2/3 times 21/64 = 7/32. Check: for 2^4:3 this argument gives 32 x 15 / 48 x 48 = 2/3 x 5/16 = 5/24. I presume 7/32 beats 5/24, but don't have time to check. Obviously 2^2n:3 will get better and better, and might be the full answer, though I'd like to check things like 2^4n : 5. JHC

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

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