TITLE: $K$ is a normal subgroup of a finite group $G$ and $S$ is a Sylow $p$-subgroup of $G$. Prove that $K \cap S$ is a Sylow $p$-subgroup of $K$.
QUESTION [6 upvotes]: (This is my first post here, so please excuse me if I'm not following proper etiquette.)
First, I noted that both $K$ and $S$ are subgroups, thus their intersection is a subgroup. We can write that $|G| = p^km$ and $|S| = p^k$ where $p$ is a prime and $m$ has no factors of $p$. By Lagrange's theorem, since the intersection is a subgroup of $S$, then the cardinality of the intersection must be of the form $p^l$ where $l \leq k$. We can write $K$ as $p^rn$ where $r \leq k$ and $n$ has no factors of $p$. If I can show that $l=r$, then it seems that would be enough to show that the intersection is a sylow $p$-subgroup of $K$, but I can't seem to get anywhere else with anything.
I recall a theorem that states that since $K$ is normal, then the intersection of $K$ and $S$ is normal in $S$ but I don't know if that helps me much. There is the corollary of Sylow's 3rd theorem in my book that states that "A Sylow $p$-subgroup of a finite group $G$ is a normal subgroup of $G$ if and only if it is the only Sylow $p$-subgroup of $G$" which seems like I could use that somehow, but I'm not too sure.

REPLY [3 votes]: Let $N$ be a Sylow $p$-subgroup of $K$. $N$ is a $p$-subgroup of $G$ such that $N \leq xSx^{-1}$ for some $ x\in G$. Then $N \leq xSx^{-1} \cap K=xSx^{-1} \cap xKx^{-1}$ since we have that $K$ is normal. Thus, $N \leq x(S \cap K)x^{-1}$, and consequently $x^{-1}Nx \leq (S \cap K)$. Now $x^{-1}Nx$ is a subgroup of $K$, and has maximal $p$-power order in $K$, since $\left|x^{-1}Nx\right|= \left|N\right|$. Since $S \cap K$ is a $p$-subgroup of $K$, $S \cap K$ is a maximal $p$-power subgroup of $K$, and hence a Sylow $p$-subgroup.