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Let $G$ be a group of order $p^nq$ where $p,q$ are distinct prime numbers and $n$ is positive integer. Suppose $G$ has at least two distinct Sylow $p$-subgroup $H'$ and $K'$ such that $H'\cap K'\neq 1$. Let $H$ and $K$ be two distinct Sylow $p$-subgroups of $G$ such that $|H\cap K|$ is maximal among all possible intersections of two distinct Sylow $p$-subgroups of $G$. Show that $q\mid |N_G(H\cap K)|$.

I tried to use the fact that normalizer of $p$-group grows but don't know how to use this here. Any hint or idea for this?

one potato two potato
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    See also https://math.stackexchange.com/questions/1804697/n-g-p-cap-q-has-more-than-one-sylow-p-subgroup-under-some-conditions?rq=1 – Nicky Hekster Jun 01 '22 at 16:34

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First thing to note is that for a group $G$ such that $|G|=p^aq$, we have $n_p\in\{1,q\}$. If you can show that $N_G(H\cap K)$ contains at least $2$ distinct Sylow $p$-subgroups (Sylow $p$-subgroups of normalizer, not of the whole group) it will follow that there are $q$ of them, so $|N_G(H\cap K)|=p^lq$.

So how one can show that there are at least $2$ Sylow $p$-subgroups? Are you familiar with a concept of a tame intersection? If not, try to show that $N_H(H\cap K)$ and $N_K(H\cap K)$ are Sylow $p$-subgroups of $N_G(H\cap K)$.

pansh1n
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