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.
 A: 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.


*

*King, Bruce W.
"Presentations of metacyclic groups."
Bull. Austral. Math. Soc. 8 (1973), 103–131. 
MR323893
DOI:10.1017/S0004972700045500

*Liedahl, Steven.
"Enumeration of metacyclic p-groups."
J. Algebra 186 (1996), no. 2, 436–446.
MR1423270
DOI:10.1006/jabr.1996.0381
A: 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.
