I thought the definition of Sylow-p-subgroup is a subgroup of maximal $p$ power, i.e. maximal $k$ such that $p^k \Big | |G|$. In your question it seems to me there are higher powers of $5$ that divide $|S_{16}| = 16!$. Can someone point out to me what I'm missing? Thanks! – 2011-02-16
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@Matt: where is the problem? the 5-sylow groups here will be of order $5^3$, cause 16! = ..5..10..15.. – 2011-02-16
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my bad, I read your question as if you were looking for subgroups of order $5$. – 2011-02-16
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Does conjugation work here? In this case we just need to find a 5-sylow subgroup of $S_{16}$ which is not normal which will fulfill the condition, don't we? If I miss something, please tell me, thanks. By the way, to type $S_{16}$ is better to use S_{16} instead of S_16 as far as I am concerned. – 2011-02-16