Suppose $ G = G_{1}G_{2} $. For any prime $ p $, there exist $ P , P_{1} , P_{2} $ such that $ P = P_{1}P_{2} $ Suppose $ G = G_{1}G_{2} $. For any prime $ p $, there exist $ P , P_{1} , P_{2} $ such that $ P = P_{1}P_{2} $, where $ P \in Syl_{p}(G) $ and $ P_{i} \in Syl_{p}(G_{i}) $ , $ i = 1,2 $.
The proof for this lemma by Huppert is written in German. Can someone help me understand it, or introduce another source?
 A: I change the notation for clarity. Let $G=HK$, $H$, $K$ subgroups and let $p$ a prime dividing the order of $G$. Then there exists a $P \in Syl_p(G)$ such that $P=(P\cap H)(P \cap K)$, with $P \cap H \in Syl_p(H)$ and $P \cap K \in Syl_p(K)$.
Proof Let us first find a Sylow $p$-subgroup $P$ of $G$ such that $P\cap H$ is a Sylow $p$-subgroup of $H$ and $P\cap K$ is a Sylow $p$-subgroup of $K$. Let $Q$ be a Sylow $p$-subgroup of $H$ and let $R$ be a Sylow $p$-subgroup of $K$. Choose a Sylow $p$-subgroup $S$ of $G$ such that $Q\subseteq S$. By Sylow theory, there is a $g\in G$ such that $R\subseteq S^g$. In particular, $S\cap H=Q$ and $S^g\cap K=R$. But $g=hk$ for some $h\in H$ and $k\in K$. Then $S^g\cap K=R=S^{hk} \cap K=(S^h \cap K)^k$, hence $R^{k^{-1}}=S^h \cap K$ and this is a Sylow $p$-subgroup of $K$, being a conjugate of $R$. On the other hand, $S^h \cap H=(S \cap H)^h=Q^h \in Syl_p(H)$, since it is a conjugate of $Q$. So $P=S^h$ is the Sylow $p$-subgroup we were looking for.
Finally we use a counting argument to show that indeed $(P \cap H)(P \cap K)=P$. Observe that $|(P\cap H)(P\cap K)|=|P\cap H||P\cap K|/|P\cap H\cap K|$. But then $|G|.\color{darkblue}{(| H \cap K|/|H \cap K \cap P|)}= \color{darkblue}{(|H|/|P \cap H|).((|K|/|P \cap K|)}.|(P \cap H)(P \cap K)|$, where the numbers in $\color{darkblue}{darkblue}$ are not divisible by $p$. In other words, $|(P\cap H)(P \cap K)|$ is divided by the largest $p$-power ($|P|$), dividing the order of $G$. And of course, the subset $(P\cap H)(P \cap K) \subseteq P$.$\square$
