About Sylow's theorem

Question

Let $ G = MN $ and let $ p $ a prime dividing the order of $ G $. Assume $ P $ is a sylow $ p $-subgroup of $ G $. Let $ H $ be a Sylow $ p $-subgroup of $ M $ and let $ K $ be a Sylow $ p $-subgroup of $ N $. Choose a Sylow $ p $-subgroup $ S $ of $ G $ that $ H \subseteq S $. Why there is a $ g \in G $ that $ K \subseteq S^{g} $?

Answer 1

$K$ is a $p$-subgroup of $N$, hence a $p$-subgroup of $G$. It is then contained in a $p$-Sylow subgroup $R$ of $G$. and from Sylow's theorem, we know that $R$ and $S$ are conjugated in $G$.