# On a finite nilpotent group with a cyclic normal subgroup

I'm reading Dummit & Foote, Sec. 6.1.

My question is the following. If $G$ is a finite nilpotent group with a cyclic normal subgroup $N$ such that $G/N$ is also cyclic, when is $G$ abelian?

I know that dihedral groups are not abelian, and I think the question is equivalent to every Sylow subgroup being abelian.

EDIT: So, the real question is about finding all metacyclic finite p-groups that are not abelian.

Thanks.

• Could you please make the title more informative? – Rasmus Oct 21 '11 at 11:48

## 2 Answers

Rod, you are right when you say this can be brought back to every Sylow group being abelian. Since $G$ is nilpotent you can reduce to $G$ being a $p$-group. However, a counterexample is easily found, take the quaternion group $G$ of order 8, generated by $i$ and $j$ as usual. Let $N$ be the subgroup $<i>$ of index 2. $N$ satisfies your conditions but $G$ is not abelian.

A finite nilpotent group is abelian if and only if all of its Sylow subgroups are abelian. This is just because a finite nilpotent group is the direct product of its Sylows.

Groups G with normal subgroups N such that both N and G/N are cyclic are called metacyclic groups.

Metacyclic p-groups need not be abelian: take the Sylow p-subgroup of the group of affine functions on $R=\mathbb{Z}/p^2\mathbb{Z}$, that is all functions $x \mapsto ax+b$ where a, b are in R and where the multiplicative order of a is a power of p. This is a group of order p3 called the extra special group of exponent p2 and order p3. For p=2 it is also known as the dihedral group of order 8.

Metacyclic p-groups were classified quite a few times. My favorites are King (1973) and Liedahl (1996). Magma has a slightly different classification available as a database, and it was easy to write code for King and Liedahl in GAP.