Help proving that a simple group of order 360 must be isomorphic to A_6
billdeja_at_weacca.com
Date: 03/16/05
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Date: 16 Mar 2005 04:54:29 0800
I am looking for a reasonably elementary proof that a simple group of
order 360 must have order 360. I have proven all sorts of things
about the Sylow subgroups, their normalizers, and their intersections;
but I haven't stumbled across anything that gives me the result.
I suspect that it might be fruitful to take a Sylow 5 subgroup P. Its
normalizer contains 5 elements of order 2. Take one of them. It
also belongs to a subgroup H of G isomorphic to A_4. If we could show
that PH is not all of G, we would be done.
Another approach might be to somehow use the fact that if H acts on G's
Sylow 5 subgroups by conjugation then Orbit(P) is of order 6, most
plausibly by somehow using this fact to partition the Sylow 5 subgroups
into 6 sets of 6, and then having G somehow act on these. Each
approach is believable because it's easy to make it work in A_6 itself.
But I cannot see how to get either of them to work.
Any hints would be appreciated  either on how to make one of these
ideras work out  or on some other approach.
BTW, the subgroups of order 12 are interesting. Some of them clearly
correspond to two pint stabilizers, e.g. the one generated by (123) and
(12)(34) clearly stabilizes 4 and 5; but I don't know what to make of
the other ones, e.g. the one generated by
(12)(34) and (135)(246).
Bill.
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 Reply: Jim Heckman: "Re: Help proving that a simple group of order 360 must be isomorphic to A_6"
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