Let $p\mid n$, then by $n_p$ we mean the $p$-part of $n$, i.e. $n_p = p^k$ if $p^k\mid n$ but $p^{k+1}\nmid n$. Let $G$ be a finite group, $M\leq G$ and $P\in Syl_p(G)$. Is It true that $|M\cap P|=|M|_p$? Why?
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1$\begingroup$ What if $M$ is a $p$-group that is not contained in $P$? $\endgroup$– Jason StarrCommented Aug 17, 2015 at 12:47
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$\begingroup$ @Jason: In this case we can consider a Sylow $p$-subgroup such that contains $M$ and so this equality is true. So we should consider $M$ is not a $p$-group. What happen in this case? This equality is true again? $\endgroup$– ninaCommented Aug 17, 2015 at 12:59
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$\begingroup$ Are you asking whether every Sylow $p$-subgroup of $M$ is contained in some Sylow $p$-subgroup of $G$? Since every $p$-subgroup of $G$ is contained in a maximal $p$-subgroup of $G$, this is true. You can find the statement of the Sylow theorems in algebra textbooks; MO is not the place for this type of question. $\endgroup$– Jason StarrCommented Aug 17, 2015 at 13:05
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$\begingroup$ As others have remarked, this is false in general. However, it is true if $M \lhd G$, or more generally, if $M \lhd \lhd G$. $\endgroup$– Geoff RobinsonCommented Aug 17, 2015 at 13:37
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$\begingroup$ Let a Sylow p-subgroup is not normal in a p-group G, P and P_1 are distinct Sylow p-subgroups of G. If M is the normalizer of P_1 in G, then |M\cap P|_p<|P| $\endgroup$– yakovCommented Jul 19, 2016 at 11:44
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The answer is NO. The easiest example is given by $G=\mathcal{S}_3$, the symmetric group on three elements. If you take $P=\langle (12)\rangle$, $M=\langle (13)\rangle$ and $p=2$, then $\vert M\cap P\vert_p=1$ but $\vert M\vert_p=2$.
answered Aug 17, 2015 at 13:25