Sylow $p$-subgroup of a finite group I know:
Let $P$ be a Sylow $p$-subgroup of a finite group $G$.If $N$ is normal in $G$, then $P \cap N$ is a Sylow $p$-subgrup of $N$.
But if $N$ is not normal in $G$ , there is also the issue?
thanks for your help.
 A: I apologize that I overlooked a detail, but here's an emendation for a nontrivial intersection:
Consider $G = S_5$, the group of permutations on the set $\{1,2,3,4,5\}$. Consider the unique Sylow $2$-subgroup (isomorphic to $D_8$), that is
$P = \langle (1234), (13) \rangle = \{1, (1234), (13)(24), (1432), (13), (24), (12)(34), (14)(23)\}$.
and the (not normal) Klein subgroup $N = \{1, (12), (34), (12)(34)\}$. Then $P \cap N = \{1, (12)(34)\}$ is not a Sylow $2$-subgroup of $N$.
A: Here is a simple example. Consider the symmetric group $\Sigma_3$ and $ P = \{ 1, (1,2) \}$, which is a $2$-Sylow. Now consider the subgroup $N = \{ 1, (1,3) \}$. Then $ P \cap N = \{ 1 \}$, which is not a $2$-Sylow of $N$.
A: If $H \leq G$ (not necessarily normal), then it works only one way: $P \in Syl_p(H)$, then $P=H \cap S$ for some $S \in Syl_p(G)$. Proof: By Sylow theory, the $p$-subgroup $P$ (seen as subgroup of $G$) is contained in some Sylow $p$-subgroup of $G$, say $S$. Hence $P \subseteq H \cap S$, and in particular $|P| \leq |H \cap S|$. But $S \cap H$ is a $p$-subgroup of $H$ and again by Sylow theory (in $H$) contained in some conjugate of $P$, say $P^h$, with $h \in H$: $H \cap S \subseteq P^h$. Hence $|H \cap S| \leq |P^h|=|P|$. So, $|H \cap S|=|P|$ and we must have $P=H \cap S$.
