I am aware of No simple group of order $300$. It is said that "there would be 6 Sylow 5-groups, one of which will have an index of 6", but why does one have index 6? If we write $|G|=p^k m$ where $p\nmid m$, then by Lagrange's theorem any $p$-Sylowgroup has index $m$, which in this case is 12?
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Indeed, that is an error (note that it does not appear in the accepted answer). – Tobias Kildetoft Jul 09 '13 at 11:37
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What you said is said in that link is a serious blunder: as all the Sylow $\,5-$subgroups obviously have the same order then they all have the same index.
What is true is that the number of different Sylow $\,5-$subgroups equals the index of the normalizer of any of them, so we have a subgroup $\,N\le G\,$ with $\,[G:N]=6\;$. This already leads to a contradiction if we suppose $\,G\;$ is simple, as then the regular action of $\,G\,$ on $\,N\,$ gives us an injective homomorphism $\,G\to S_6\,$...but this is impossible (why?) .
DonAntonio
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1Since in this case $G$ is isomorphic to a subgroup of $S_6$, but the order of $G$ does not divide the order of $S_6$. Thanks for clarification! – Jul 09 '13 at 12:01