Existence of $A \in \operatorname{Syl}_p(H)$ and $B \in \operatorname{Syl}_p(K)$ such that $AB \in \operatorname{Syl}_p(G)$ $\DeclareMathOperator{\Syl}{Syl}$Assume that $G = HK$ is a finite group, where $H$ and $K$ are two subgroups of $G$. I want to find an $A \in \Syl_p(H)$ and $B \in \Syl_p(K)$ such that $AB \in \Syl_p(G)$, where $p$ is prime.
When $p$ divides $|H|$ and does not divide $|K|$, there exist $A \in \Syl_p(H)$ and $B=\{e\} \in \Syl_p(K)$ such that $AB = A \in \Syl_p(G)$. However, I have no idea for the case that $p$ divides both $|H|$ and $|K|$.
 A: 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 $$|(P \cap H)(P \cap K)|=\frac{|P \cap H| \cdot |P \cap K|}{|P \cap H \cap K|}=\frac{|H|_p \cdot |K|_p}{|P \cap H \cap K|}$$
where the $p$-subscript denotes the largest $p$-power dividing a positive integer (which is understood to be $1$ if the integer in question is not divisible by $p$).
Since $P \cap H \cap K$ is a $p$-subgroup of $H \cap K$, note that $|P \cap H \cap K| \leq |H \cap K|_p$. Combining this:
$$|(P \cap H)(P \cap K)| \geq \frac{|H|_p \cdot |K|_p}{|H \cap K|_p}=[\frac{|H| \cdot |K|}{|H \cap K|}]_p=|G|_p=|P|$$ since $G=HK$ and $P \in Syl_p(G)$. As a set $(P \cap H)(P \cap K) \subseteq P$, so we conclude $P=(P \cap H)(P \cap K)$.$\square$
