Give an example of a non-abelian group G containing a proper normal subgroup N such that $G/N$ is abelian.

Question

Give an example of a non-abelian group G containing a proper normal subgroup N such that $G/N$ is abelian.

I KNOW THERE IS A QUESTION OF THE SAME NAME.

However, I need more involved assistance.

My professor suggested looking at the dihedral group $D_6$. I intend to find a homomorphism with a kernel that is the subgroup. The kernel of this homomorphism is a subgroup of G and is obviously normal.

However, why does this imply that $G/N$ is abelian, esp. since G itself is non-abelian?

Answer 1

HintL: If the size of $G/N$ is very small, that will force Abelian. One can also find an example using as $G$ a suitable symmetric group.

Answer 2

Hint: Let $N$ be nonabelian, let $A$ be abelian non-trivial. Consider $N\times A$ and the projection $\pi\colon N\times A\to A$.