Sylow subgroup of a product of subgroups Let $UV$ be a subgroup of $G$ and $P$ is a Sylow $p$-subgroup of $G$. Suppose that $P \cap U \in \operatorname{Syl}_p(U)$ and $P \cap V \in \operatorname{Syl}_p(V)$. I need to show that $P\cap UV \in \operatorname{Syl}_p(UV)$
$P \cap UV$ is clearly a $p$-subgroup of $UV$. Also $(P \cap U)(P \cap V) \subseteq P \cap UV$. We need to show that it is a maximal $p$-subgroup of $UV$ but I'm not sure how to proceed
 A: We use a counting argument to show that indeed $(P \cap U)(P \cap V)=P \cap UV$. Observe that $|(P\cap U)(P\cap V)|=\frac{|P\cap U| \cdot|P\cap V|}{|P\cap U\cap V|}=\frac{|U|_p \cdot |V|_p}{|P \cap U \cap V|}$, where the index $p$ denotes the largest power of $p$, dividing the order of the group between its $|\cdot|$. Now, $|P \cap U \cap V| \leq |U \cap V|_p$, since $P \cap U \cap V$ is a $p$-subgroup of $U \cap V$.
Hence, $$|(P\cap U)(P\cap V)| \geq \frac{|U|_p \cdot |V|_p}{|U \cap V|_p}=\left(\frac{|U| \cdot |V|}{|U \cap V|}\right)_p=|UV|_p.$$
Since as a set $(P \cap U)(P \cap V) \subseteq P \cap UV$, and $|P \cap UV| \leq |UV|_p$, this forces $(P \cap U)(P \cap V)=P \cap UV$, and $P \cap UV \in Syl_p(UV)$.
A: The OP is supposing that there is a $P \in Syl_p(G)$ with $P \cap U \in Syl_p(U)$ and $P \cap V \in Syl_p(V)$. We will show that this is in fact always guaranteed.
Let us first find a Sylow $p$-subgroup $S$ of $UV$ such that $S\cap U$ is a Sylow $p$-subgroup of $U$ and $S\cap V$ is a Sylow $p$-subgroup of $V$. 
Let $Q$ be a Sylow $p$-subgroup of $U$ and let $R$ be a Sylow $p$-subgroup of $V$. Choose a Sylow $p$-subgroup $T$ of $UV$ such that $Q\subseteq T$. By Sylow theory, there is a $g\in UV$ such that $R\subseteq T^g$. In particular, $T\cap U=Q$ and $T^g\cap V=R$. But $g=uv$ for some $u\in U$ and $v\in V$. Then $T^g\cap V=R=T^{uv} \cap V=(T^u \cap V)^v$, hence $R^{v^{-1}}=T^u \cap V$ and this is a Sylow $p$-subgroup of $V$, being a conjugate of $R$. On the other hand, $T^u \cap U=(T \cap U)^u=Q^u \in Syl_p(U)$, since it is a conjugate of $Q$. So $S=T^u$ is the Sylow $p$-subgroup we are looking for.
Now, by Sylow theory applied to subgroups, $S=P \cap UV$ for some $P \in Syl_p(G)$. Hence $P \cap U=P \cap UV \cap U=S \cap U \in Syl_p(U)$, and similarly $P \cap V \in Syl_p(V)$.
