True or not: If a normal subgroup and its quotient are commutative, then the group is.

Question

Let $G,*$ be a group and $A,*$ a normal subgroup of $G,*$: $$ A \triangleleft G \quad( \equiv \forall g\in G: gA = Ag) $$ Then $G$ is commutative iff A and $\frac{G}{A}$ are commutative.

I can see that the theorem is true in one direction $(\Leftarrow)$, but I'm not sure about the other. Any hints? Or maybe a counterexample...

Answer 1

This is false. $A_3$ and $S_3/A_3$ are both abelian, but $S_3$ is not.

Answer 2

Check $S_3$ ${}{}{}{}{}{}{}{}{}{}{}$

Answer 3

The commutator $G'$, subgroup generated by all $aba^{-1}b^{-1}$, is normal and also $G/G'$ is abelian in any group. So, take any non-abelian group to illustrate that your hypothesis aren't enough, meanwhile $G'$ is non-trivial, i.e. $1<|G'|$ and $G'\neq G$.