Let $G$ be a group and $H,K,L$ be subgroups with $H \subseteq K$. Prove that $K \cap HL=H(K\cap L)$ 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??
 A: We show directly double set contention:
$$\begin{align}&x\in K\cap HL\implies x=k=hl\implies l=h^{-1}k\in HK=K\implies\\{}\\& \implies l\in K\cap L\implies x=hl\in H(K\cap L)\implies \color{red}{K\cap HL\subset H(K\cap L)}\\{}\\&\text{ and on the other side}\\{}\\
&x\in H(K\cap L)\implies x=ht\,,\,\,h\in H\le K\;,\;\;t\in K\cap L\le \begin{cases}K\\L\end{cases}\implies\\
&\implies ht\in KK=K\;,\;\;ht\in HL\implies ht\in K\cap HL\implies\color{red}{H(K\cap L)\subset K\cap HL}\end{align}$$
A: First of all, $HK=\{hk\mid h\in H,k\in K\}$ is the set product or complex product (it need not be a group itself), similarly for the other set products. 
Since $H\subseteq K$ and both are groups we have $HK=K$. Now we have $K\cap L\subseteq K$, hence $H(K\cap L)\subseteq HK=K$; analogously, $K\cap L\subseteq L$ implies $H(K\cap L)\subseteq HL$. Altogether $H(K\cap L)\subseteq HK\cap HL = K\cap HL$ (in a sense, this is the trivial inclusion). 
For the other inclusion take any $x\in K\cap HL$; that is, $x=hl$ for some $h\in H, l\in L$ (note $x\in K$); then $l=h^{-1}x\in HK=K$, thus $l\in K\cap L$ and $x=hl\in H(K\cap L)$.
A: 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
