Prove a quotient group is abelian

Question

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

Answer 1

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.

Answer 2

$$G/M\;\;\text{abelian}\;\iff\;G'=[G:G]\le M$$

But

$$M\le N\implies\;\text{also}\;\;G'\le N\implies G/N\;\;\text{abelian}$$

Answer 3

Note that $xNyN:=xyN$ and $xMyM:=xyM$. The fact that $N$ and $M$ are normal implies that these multiplications are well defined.

It is to be shown that for each pair $x,y\in G$ we have $xyN=yxN$ or equivalently $xyx^{-1}y^{-1}\in N$. This is a direct consequence of $xyx^{-1}y^{-1}\in M\subseteq N$ (equivalent with $xyM=yxM$).

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The commutatorsubgroup of $G$, often denoted as $[G,G]$, is the subgroup that is generated by elements of the form $xyx^{-1}y^{-1}$. It is a characteristic, hence normal subgroup and is somehow decisive when it comes to the question whether a quotient $G/K$ is abelian:

$$G/K\text{ is abelian }\iff [G,G]\subseteq K$$