## If the quotients of a group by two subgroups are abelian, then the quotient by their intersection is abelian

Let $A,B \leq G$ be normal subgroups such that $G/A$ and $G/B$ are abelian. Prove that $G/(A \cap B)$ is abelian.

Since $G/A$ is abelian we have $[G,G] \leq A$. Likewise, $[G,G] \leq B$. Thus $[G,G] \leq A \cap B$, so that $G/(A \cap B)$ is abelian.