# How to reach $Syl_p(H)\neq Syl_p(G)$?

I am really trapped to tis problem: If $G$ is a simple finite non-abelian group then for every proper subgroup $H$ and prime $p$ ($p||G|$) we have $Syl_p(H)\neq Syl_p(G)$. I thank for any help.

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If ${\rm Syl}_p(H) = {\rm Syl}_p(G)$, then the subgroup of $G$ generated by all of the Sylow $p$-subgroups of $G$ is contained in $H$, and is therefore a proper subgroup of $G$. What other property does that subgroup have? – Derek Holt Sep 19 '12 at 10:39
The above comment is practically the whole answer: why not write it as such instead of "comment"? – DonAntonio Sep 19 '12 at 11:11

Consider $O^{p'}(X) = \langle P : P \in \newcommand{\Syl}{\operatorname{Syl}}\Syl_p(X) \rangle$ to be the subgroup generated by the elements of $p$-power order. This is a normal subgroup since its generating set is closed under conjugation. If $\Syl_p(X) = \Syl_p(Y)$ then $O^{p'}(X) = O^{p'}(Y)$ (iff in fact) and obviously $O^{p'}(X) \leq X$.
Hence if $\Syl_p(H) = \Syl_p(G)$, then $O^{p'}(H) = O^{p'}(G)$ is normal in $G$ (being $O^{p'}(G)$) and proper (being $O^{p'}(H)$, which is contained in $H$, a proper subgroup of $G$). Since $G$ is simple, $O^{p'}(G)=1$, but this means that every Sylow $p$-subgroup is trivial, so that $p$ cannot divide $|G|$. You do not need that $G$ is non-abelian, though the abelian case is a bit silly.