TITLE: Nilpotent and cyclic quotient group implies abelian
QUESTION [5 upvotes]: I'm stuck on the follow question, and honestly, I cannot think of a way to deal with it. Does anyone have any suggestions?  Any hints would be very appreciated! Thanks

Suppose that $G$ is a nilpotent group and that $G/G'$ is a cyclic
  group, where $G'$ denotes the commutator subgroup of $G$. Show that
  $G$ is abelian.

Intuition:  My idea is to do something along the line of induction on the length of the upper central series that $G$ has, and showing that, if $1 =G_n \leq \dots \leq G_2=G' \leq G_1=G $, then $1=G_n = \dots = G'$.

REPLY [2 votes]: If $G$ has class $1$ or  $2$, then $G' \le Z(G)$ so $G/Z(G)$ is cyclic and $G$ is abelian.
Otherwise, we can apply induction to $G/Z(G)$ and deduce that $G/Z(G)$ is abelian, but then $G$ has class at most $2$. Note that this argument does not assume that $G$ is finite.