Can a nonabelian group $G$ have a normal abelian subgroup $H$ with $[G:H]=3$? If $H$ is a cyclic group, for instance, and $[G:H］=2$, then $G$ is a dihedral group with elements equal to the possible states of a regular polygon upon which rotations and reflections are applied. In this case, the quotient group $G/H$ of order $2$ can be interpreted as the group whose elements are the states "not reflected" and "reflected."
I'm having lots of trouble imagining what it would mean if $H$ were a cyclic normal subgroup of $G$ and $[G:H]=3$, or any odd number, for that matter. For even numbers, however, I can imagine the group whose elements are states of a regular polygon upon which rotations and reflections are applied, and whose vertices change color whenever two reflections are applied. In this case, if $H$ is the subgroup of rotations, then  $[G:H］$ could be made to be any even numbers. However, I can't imagine any such thing for odd numbers.
Any hints or advice would be greatly appreciated. For context, I'm doing research about Cayley graphs, and my group theory knowledge isn't as good as it should be. I tried looking through textbooks to find the answer to this question, but I haven't been successful.

Yes.  The Klein subgroup $\{1, (1~2)(3~4), (1~3)(2~4), (1~4)(2~3)\} \triangleleft A_4$.  This subgroup is abelian, as the title of the question requests, but it's not cyclic.
