# Help proving that a simple group of order 360 must be isomorphic to A_6

bill-deja_at_weacca.com
Date: 03/16/05

```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.

## Relevant Pages

• Re: Help proving that a simple group of order 360 must be isomorphic to A_6
... >I suspect that it might be fruitful to take a Sylow 5 subgroup P. ... >plausibly by somehow using this fact to partition the Sylow 5 subgroups ... This gets you a permutation action on 10 points. ...
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