# If $G$ nilpotent and $G/G'$ is cyclic then $G'=1$.

Hi: This question has already been answered here: Nilpotent group such that $G$/$G'$ is cyclic or a Prüfer group

But I do not understand the proof given in the answer.

$$G$$ is a nilpotent group and $$G/G'$$ is cyclic. Prove that $$G'=1$$. That is, I must prove that $$G$$ is abelian.

If I had $$G' \le Z(G)$$ then by a theorem I would have $$G$$ abelian.

But $$G'$$ (the derived group of $$G$$) has not to be in the center of $$G$$.

The author defines nilpotency using the upper central series. He doesn't even mention central series in general.

Suppose the length of the upper central series for $$G$$ is $$n$$. $$n=2$$: $$1 \lt Z(G) \lt G$$. By another book I know the length of the lower central series is two too: $$1 \lt G' \lt G$$ and also that $$G' \le Z(G)$$. Then $$G/G'$$cyclic gives $$G$$ abelian. $$n=3$$: $$1 \lt Z_1 \lt Z_2 \lt G$$. Here $$Z_1=Z(G), Z_2/Z_1=Z(G/Z_1)$$ and $$G/Z_2$$ is abelian.

The lower central series is $$1 \lt G_3 \lt G_2 \lt G$$ with $$G_3 \le Z_1$$. That is I no longer have $$G' \le Z_1$$. There must be some way of using the case $$n=2$$ here. But I don't find it.

Could you give me a hint?

Let $$1 = Z_0 \lt Z_1 \lt\cdots \lt Z_{n-1} \lt Z_n = G$$ be the upper central series of $$G$$.
Assume that $$n\geq 2$$.
Since $$G/Z_{n-1}$$ is abelian, it follows that $$G'\leq Z_{n-1}$$. In particular, $$G/Z_{n-1}$$ is a quotient of $$G/G'$$, and hence cyclic. But $$Z_{n-1}/Z_{n-2}$$ is the center of $$G/Z_{n-2}$$, and $$G/Z_{n-1} \cong (G/Z_{n-2})/(Z_{n-1}/Z_{n-2})$$. That is, $$G/Z_{n-2}$$ is such that if you mod out by its center, you get a cyclic group. This means that $$G/Z_{n-2}$$ is abelian. But that would mean that $$Z(G/Z_{n-2}) = G/Z_{n-2}$$, and hence that $$Z_{n-1}=G$$.
But this is impossible. The contradiction arises from the assumption that $$n\geq 2$$, and therefore $$n\leq 1$$; that is, $$G$$ is abelian.