# 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.

• Here's a simple infinite example: take $\mathbb Z^3 \rtimes (\mathbb Z/3\mathbb Z)$, where $t$ is a generator of the $\mathbb Z/3\mathbb Z$ factor and conjugation by $t$ cyclically permutes the factors of $\mathbb Z^3$, i.e. $t^{-1}(a,b,c)t = (c,a,b)$ for $(a,b,c) \in \mathbb Z^3$. $\mathbb Z^3$ is normal and has index $3$. – Rylee Lyman Apr 26 '19 at 1:35

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.
More generally, let $$p$$ be any prime. If one has a finite group $$G$$ of order $$p^n$$, then it has a composition series with quotients of order $$p$$, that is there is a sequence $${e}=G_0\triangleleft G_1\triangleleft G_2\triangleleft \cdots \triangleleft G_{n-1} \triangleleft G_n=G$$ with each $$|G_j|=p^j$$. Each $$|G_j:G_{j-1}|=p$$. If $$G$$ is non-Abelian, then somewhere the sequence flips from Abelian to non-Abelian, so there is some $$j$$ with $$G_{j-1}$$ Abelian, and normal of index $$p$$ in the non-Abelian group $$G_j$$.
Here's a construction for the original problem with $$H$$ cyclic. Let $$H$$ be a cyclic group of prime order $$p$$ with $$p\equiv 1\pmod3$$ (for instance $$p=7$$). Then $$H$$ has an automorphism of order $$3$$. Now let $$G$$ be the semidirect product of $$H$$ by a cyclic group of order $$3$$ acting on $$H$$ via this automorphism.
This construction also works for other values of $$3$$.....