We know that the following fact holds: If $G$ is finite group and $N$ is normal in $G$ and $P\in \text{Syl}_p(G)$ then $P\cap N\in \text{Syl}_p(N)$. Also converse is also true that if $Q\in \text{Syl}_p(N)$ then $Q=P\cap N$ where $P\in \text{Syl}_p(G)$.
But I was wondering what if we omit the condition of normality? Can we conclude that $P\cap N$ will be Sylow $p$-subgroup of $N$?
An easy counterexample is given by any Sylow subgroup that isn't normal. Its intersection with another Sylow subgroup is not a Sylow subgroup of itself.