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.
How to reach $Syl_p(H)\neq Syl_p(G)$?
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group-theory
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2If ${\rm Syl}_p(H) = {\rm Syl}_p(G)$, 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? – 2012-09-19
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0The above comment is practically the whole answer: why not write it as such instead of "comment"? – 2012-09-19