# normalizer of a Hall subgroup in another Hall subgroup

Let $H$ be a Hall subgroup in a finite group G with $|H|=n$ (namely $gcd(n,|G:H|)=1$). Let $K$ be a subgroup of G with $|K|=n$, prove that $N_K(H)=H \cap K$. Moreover, show that $H$ is the unique subgroup of $N_G(H)$ with order $n$ .

Note: For the first question it's clear that $H \cap K \leq N_K(H)$, I attempted to show $|H \cap K| = |N_G(H) \cap K|$ by applying the formula $|AB|=|A||B|/|A \cap B|$ however I counld not proceed any further. For the second question I know if n is prime then by Sylow conjugate theorem one can show H is the unique sylow-n subgroup of $N_G(H)$, but I do not know how to prove for the case where n is composite.

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Here is an argument for the second part of the question: Let $T$ be another subgroup of $\mathbf{N}_G(H)$ with order $n$. As $T \subseteq \mathbf{N}_G(H)$, $TH$ is a subgroup. Since $|TH| = \frac{|T||H|}{|T \cap H|}$, every prime divisor of $|TH|$ divides $n = |H|$. Therefore $H$ being a Hall subgroup forces that the inclustion $H \subseteq TH$ can't be strict; thus $T \subseteq H$; comparing the orders we get $T = H$.
Let $\pi$ be the set of primes that divide $n$. Note that any $\pi$-subgroup (a subgroup whose order is divisible only by primes in $\pi$) of $G$ must have order dividing $n$.
Let $M=N_K(H)$. Then $HM=MH$, so $HM$ is a subgroup. Moreover, as you note, we have $|HM| = |H||M|/|H\cap M|$. Since $|M|$ divides $|K|=n$, then $|HM|$ is a $\pi$-subgroup of $G$, hence has order dividing $n=|H|$. Thereforem, $|M|/|H\cap M| = 1$, so $N_K(H) = M = H\cap M\subseteq H\cap K\subseteq N_K(H)$, giving the desired equality.