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?


1 Answer 1


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.

  • $\begingroup$ Beautiful proof. Thanks Arturo. $\endgroup$
    – stf91
    Commented Oct 27, 2019 at 23:35

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