Re: [math-fun] non-square products of squares?
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? Thanks for explaining this stuff. I'm really in the dark about being able to describe the group I'm talking about. Dan
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
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Dan Hoey -
John Conway