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I follow a proof which show that any group of order $36$ has a normal Sylow subgroup.

The author suppose that $G$ has no normal subgroups of order $9$ or $4$. Then $G$ won't have subgroups of order $18$. We take 2 subgroups of order $9$, say $P$ and $Q$, and note that their intersection is a subgroup $H$ of order $3$ which is normal in both $P$ and $Q$ and hence normal in $G$. Then $G/H$ has a normal subgroup of order $4$ and $G$ has a normal subgroup $M$ of order $12$ containing $H$. Let $S$ be a subgroup of order $4$ contained in $M$. Then $N_G(S)\not=S$ or $M$, and the proof continues. It was clear for me that $N_G(S)\not=M$ but I need some hints to understand why $N_G(S)\not=S$.

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Hint: if $N_G(S)=S$ then $S$ has $|G:N_G(S)|=9$ different conjugates and since $M$ is normal, all of them must lies in $M$. Can it be possible ?

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