How to show that group of order 760 is not simple?
By Sylow's theorem $n_{19}=20$, and $o(N(P)) = 38$, but how to continue after this?
Thanks for any help
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How many 5-Sylow subgroups can there be? – Steve Kass Mar 24 '14 at 04:48
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possible duplicate of Question on Simple Groups – Jack Schmidt Mar 24 '14 at 17:07
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Also see http://math.stackexchange.com/questions/370731/groups-with-20-sylow-subgroups for a follow up – Jack Schmidt Mar 24 '14 at 17:08
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I really cannot understand why people are voting to close this question! In the permutation action of the group by conjugation on its set of $20$ Sylow $19$-subgroups, an element of order $2$ normalizing a $19$-cycle would consist of $9$ transpositions, and hence would be an odd permutation. So intersecting with the alternating group gives a normal subgroup of index $2$.
Derek Holt
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1I voted to close as a duplicate. It seems completely on topic to me, and showed more work than the original one. – Jack Schmidt Mar 24 '14 at 17:09