If $B\triangleleft A\subset G,H\triangleleft G$ and $A/B$ abelian, then $(A\cap H )/(B\cap H)$ is abelian

Question

Just a quick check of reasoning.

Let G be a group, $B\triangleleft A\subset G,H\triangleleft G$ and $A/B$ abelian, then $(A\cap H)/(B\cap H)$ is also abelian: for any $x,y\in A\cap H$ we have $[x,y]\in B$ and $[x,y]\in H$, so that $[A\cap H,A\cap H]\subset B\cap H$.

Thanks.

Answer 1

Only one thing needs to be fixed in your argument (maybe a typo?), namely $[A \cap H,A \cap H] \subseteq B \cap H$, not $\in$. When applying $[-,-]$ to elements, you get elements ; if you apply it to subsets (e.g. subgroups), then you get a subset. The rest is fine!

Hope that helps,