Re: [math-fun] non-square products of squares?
I wrote:
Up to order 255, the only group that exceeds 1/6 is SmallGroup(48,50), or
<a,b,c : a^3 = b^2 = c^2 = (a b)^3 = (a c)^3 = (b c)^2 = (a b a^-1 c)^2 = 1> .
It has a probability of 5/24 that a,b,ab are square,square,nonsquare.
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 . Of course G/G' is C3. But the direct product C3 x 2^4 is not G. That seems similar to the case for A4: A4/C3 = 2^2 but 2^2x3 is not A4. Somehow I thought that was the way to decompose and reconstitute groups, but I guess I'm mistaken. As for identifying the group, I have checked that its order spectrum [1,15,32,0,...,0] is unique. Its center is 1. Dan
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 ?
Of course G/G' is C3. But the direct product C3 x 2^4 is not G.
That seems similar to the case for A4: A4/C3 = 2^2 but 2^2x3 is not A4. Somehow I thought that was the way to decompose and reconstitute groups, but I guess I'm mistaken.
Well "this" IS the way, but of course you must get "this" right! What happens is that A4 does have a normal subgroup 2^2, but that this subgroup is not a direct factor, because the "complementary" subgroup 3 doesn't fix the 2^2 pointwise, but rather permutes its three non-identity elements cyclically. In the "ATLAS" slang it has shape 2^2:3 rather than 2^2 x 3. You might be interested to see what I regard as the best ways of thinking about all the other small groups - I went through all orders less than 32 yesterday. The hardest case, after 16, is of course 24, for which the non-abelian groups are six direct products: 2xQ12, 2xT12, 3xD8, 3xQ8, 4xD6, 2^2xD6 and six indecomposable ones D24, Q24, O24, 2T12, (12 :^5 4)/2, (4,6|2,2). Here the "tetrahedral group" T12 is another name for A(4) = 2^2:3 and the "octahedral group" O24 another name for S(4) = 2^2:D6. The group 2T12 could alternatively be described as Q8:3. The next-to-last group here is obtained from 12 :^5 4 = < a,b | 1 = a^12 = b^4, b^-1.a.b = a^5 > by identifying the central subgroups 2 of its "12" and "4", so that it has the poresentation (12 :^5 4)/2 = < a,b | above with a^6 = b^2 >. In general the group (m,n|2,2) is defined by the presentation < a,b | 1 = a^m = b^n = (ab)^2 = (a.b^-1)^2 > and this is only a sensible name when m and n are both even. Structurally, it has a normal subgroup generated by a^2 and b^2 that is the direct product of cyclic groups of orders m/2 and n/2, but I can't really find a good "structural notation" that conveys how this completes to the entire group. JHC
In my recent response to Allan, I should of course have said "not just doubt, but moral certainty of failure". After some more reflection, I've come to the conclusion that Allan's "Tightness Conjecture" probably WILL fail in 3 dimensions as well as in 4, and that Jud can probably redeem himself (so to speak) by establishing this. Namely, among the many packing configurations that Sloane et al have produced with the "Gosset" program, there are probably several records that aren't (currently) achieved by any "tight" configuration. Now if Jud were to run his program for spherical caps of the appropriate radius (minus, for safety, a very small epsilon), it should prove this. It would also be of interest to run it for the case of 4-dimensional unit spheres, to see just how far short of 24 it falls. Let me just conjecture here that it won't even find 23. I'm sure the program would also find a few unforeseen things that would amply justify the trouble of writing it. How about it, Jud? JHC
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Dan Hoey -
John Conway