Let $G$ be a group with a normal subgroup $M$ such that $G/M$ is abelian. Let $N\geq M$ and $N \unlhd G$. Show $G/N$ is abelian.
My attempt:
To show that $G/N$ is abelian, we need to show that for all $x,y \in N,~ xNyN=yNxN$
$G/M$ is abelian, so for any $x,y \in M, xMyM=yMxM$. Since $M$ is a subgroup of $N$, the cosets $xM$ and $yM$ are in $B$. Then, I am stuck, but I feel like I should construct some sort of algebraic manipulation to show $G/N$ is abelian.
Thanks in advance
Best Answer
By correspondence theorem $N/M$ is normal in $G/M$.
By third isomorphism theorem $$G/N \cong \frac{G/M}{N/M} $$But $G/M$ is abelian, and a quotient of an abelian group is abelian, so $G/N$ is abelian.