Example of normal subgroup that contains the commutator group as a proper subgroup.

Question

In my algebra class, we went over the theorem:

Suppose that $N$ is a normal subgroup. Then $G/N$ is abelian iff $[G:G]\subseteq N$

The proof we used seemed to imply that the normal subgroup is the commutator. I know that this observation is actually meaningless, an artifact of the exact proof we used, but my professor couldn't think up an example to the contrary, neither could wikipedia, and I would like one.

Just to be clear, by proper, I mean also non-trivial, rather than simply of a smaller cardinality.

Answer 1

Often one can find examples by looking at groups of small order. The quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$ has commutator subgroup $[Q_8,Q_8]=\{\pm 1\}$. All subgroups of order $4$ contain the commutator subgroup as a proper subgroup; namely $\{\pm 1,\pm i\}$, $\{\pm 1,\pm j\}$, and $\{\pm 1,\pm k\}$. Moreover all subgroups of order $4$ are of index $2$, and hence normal subgroups.

Answer 2

What you are wanting is a group $G$ such that $G^{ab}=G/[G, G]$ has some proper, non-trivial homomorphic image (that is, the abelianisation $G^{ab}$ is not cyclic of prime order). Can you see why this is what you need?

So knowing this, examples can be conjured up easily! For example, $S_3\times\mathbb{Z}_6$ works (why?).