Let $G$ be a group and $H,K,L$ be subgroups with $H \subseteq K$. Prove that $K \cap HL=H(K\cap L)$
Ok, so I know that $H,K,L$ are subgroups of $G$ which tells me that $H,K,L$ are groups themselves and that $H,K,L$ are also closed under the product of $G$.
So, if I have $H,K,L \lt G$ and $H \subseteq K$
Then can I say $K \cap HL= H(K \cap L)$ Therefore, $K \cap HL \subseteq H(K\cap L)$ and $H(K\cap L) \subseteq K \cap HL$ So, if I can show $a \in K \cap KL$ where $a=hl \in K$ where $h \in H$ and $l \in L$ Would this suffice??
Suppose $a \in K \cap HL \implies a \in K$ and $a\in HL \implies a=hl$ where $h \in H, l \in L$
But $K$ is a group $\implies hl \in K$ so $h,l \in K$ and also $l \in K \cap L$
Suppose $b \in K \cap L \implies b \in K$ and $b \in L$
Since $H \subseteq K \implies \forall h \in H$, $hb \in K$
Hence in both the cases we obtain the same group K