Let $\mathrm G$ be a group. $\mathrm H$ and $\mathrm K$ subgroups of $\mathrm G$ such as $\mathrm H$$\lhd$$\mathrm K$. Prove that for any subgroup $\mathrm L$ of $\mathrm G$, we have
$\mathrm H$$\cap$$\mathrm L$$\lhd$$\mathrm K$$\cap$$\mathrm L$
I've proven that $\mathrm H$$\cap$$\mathrm L$ and $\mathrm K$$\cap$$\mathrm L$ are both subgroups of $\mathrm G$ what I can't prove in mathematical writing is that $\mathrm H$$\cap$$\mathrm L$ is a ?$normal subgroup$? of $\mathrm K$$\cap$$\mathrm L$
Is this starting a way of getting there?
$\forall$n$\in$($\mathrm H$$\cap$$\mathrm L$) $\forall$m$\in$($\mathrm K$$\cap$$\mathrm L$) : $mnm^{-1}$ $\in$ ($\mathrm H$$\cap$$\mathrm L$) and $\forall$n$\in$ ($\mathrm H$$\cap$$\mathrm L$) $\forall$m$\in$($\mathrm K$$\cap$$\mathrm L$) : $mn$$m^{-1}$ $\in$ ($\mathrm K$$\cap$$\mathrm L$)
Given $x\in K\cap L$ then $xLx^{-1} = L$ because $x\in L$ while $xHx^{-1}=H$ because $x\in K$ and $K$ is contained in the normalizer of $H$. Therefore, because $g\mapsto xgx^{-1}$ defines a permutation on $G$, namely the $G$ inner automorphism induced by $x$, we must have
$$x(H\cap L)x^{-1}=xHx^{-1}\cap xLx^{-1}=H\cap L$$