On Wed, 24 Sep 2003, Dan Hoey wrote:
John Conway <conway@math.princeton.edu> wrote:
On Wed, 24 Sep 2003, Dan Hoey wrote:
I looked at your classification of groups of order 16 and verified your translation of the smaller anonymous groups I found. In hopes of identifying this group, I found its derived subgroup 2^4 .
This must be a typo for 2^2 ?
No. I'm trying to describe the 48-element group with order spectrum [1,15,32,0,...,0]. Gap claims its derived subgroup (which it defines as the subgroup generated by the commutators) is the 16-element group with order spectrum [1,15,0,0]--that's 2^4. Could GAP be mistaken, or am I just spreading confusion?
Aha. I was confused because you'd just been talking about groups of order 16, not 48. This group has shape 2^4:3 in the ATLAS slang, where the element of order 3 has no non-trivial fixed point in the 2^4 subgroup, and this uniquely determines it to be < a,b,c; d,e,f; g > subject to the relations a,b,c,d,e,f are mutually commuting elements of order 2 and abc = def = 1 and g has order 3, and transforms a -> b -> c and d -> e -> f. Here a,b,c and d,e,f are mutually commuting subgroups of order 4, each of which is extended by g to a group 2^2:3 = A(4). Since we can obtain this group from the direct product < a,b,c; g> x < d,e,f; g'> of two A(4) groups by identifying g with g', we could call it A(4)^2/3, but to my mind 2^4:3 is already a good enough description. Let me say why. The question is, is there any non-trivial element of the 2^4 that is fixed by the 3-element under conjugation? If there is, we can take it as one of the generators, and so express the group as a non-trivial direct product. But in that case, we'd use a name coming from the direct product decomposition, whatever that was. So the only group(s) for which it's sensible to express the structure as 2^4:3 will have no such fixed element. Now if a is any non-trivial element of the 2^4 we can suppose the element of order 3 - call it g - takes a -> b -> c (and so back again to a), and here we must have abc = 1, since it's fixed by g. That "pulls out" one 2^2 group, and we can repeat the argument to find a second such triple d,e,f outside <a,b,c> and so identify the group. JHC