Prove or disprove: If H is a normal subgroup of G such that H and G/H are abelian, then G is abelian.

Question

it seems like it... should be? In that I can't think of any counterexamples off the top of my head. I was looking up these http://en.wikipedia.org/wiki/Hamiltonian_group and saw the quaternion group, but the subgroups of the quaternion group aren't abelian...

Answer 1

Not true: try $G=S_3$, with $H=A_3$. By the way the following is true. A group is called metabelian if $G''=1$ (the commutator subgroup of the commutator subgroup is trivial). If $H \unlhd G$ is abelian with $G/H$ abelian, then $G' \subseteq H$ and hence $G'' \subseteq H' = {1}$. The other way around, if $G$ is metabelian then $G'$ is an abelian normal subgroup and $G/G'$ is of course abelian. So the property in the post occurs only in the class of metabelian groups. $S_3$ is an example of a non-abelian metabelian group. There are many more, like $Q$, the quaternion group of order $8$ as discussed above, or $S_3 \times Q$.