Finite extensions of nilpotent groups

Question

Let $G$ be a torsion free, nilpotent group and let $H$ be a normal subgroup of $G$ such that $G/H$ is a finite cyclic factor group. It is true that if $H$ is nilpotent of class $c$ then also $G$ has nilpotent class $c$?

How about the proof? Any suggestions?

Answer 1

An alternative proof to that of Derek Holt could be the following.

Assume $G$ has class $c$. We want to prove that $H$ has also class $c$. Clearly $H$ has class at most $c$.

There are elements $a_{1}, \dots, a_{c} \in G$ such that $$ [a_{1}, a_{2}, \dots, a_{c}] \ne 1. $$ (This a left-normed, say, commutator.)

Suppose $\lvert G / H \rvert = n$. (I am only assuming $G/H$ finite, like Derek does.) Then $a_{i}^{n} \in H$ for each $i$.

Arguing by contradiction, suppose $H$ has class less than $c$, so that we have $$ 1 = [a_{1}^{n}, a_{2}^{n}, \dots, a_{c}^{n}] = [a_{1}, a_{2}, \dots, a_{c}]^{n^{c}}. $$ (Here I have used the fact that, since $G$ has class $c$, the function $$ G \times \dots \times G \to G, \qquad (a_{1}, \dots, a_{c}) \mapsto [a_{1}, a_{2}, \dots, a_{c}] $$ is multiplicative in each argument.)

Since $G$ is torsion-free, we obtain $$ [a_{1}, a_{2}, \dots, a_{c}] = 1, $$ a contradiction.

Answer 2

I think that if $G$ is a torsion free nilpotent group of class $c$ and $G/H$ is any finite factor group, then $H$ must also have class $c$.

Suppose $G$ is generated by a set $X$, and let $X^n = \{x^n : x \in X\}$, where $|G/H|=n$. Then $X^n \subset H$, so it's enough to show that $K = \langle X^n\rangle$ has class $c$.

Let $G = \gamma^1(G) > \gamma^2(G) > \cdots$ be the lower central series of $G$.

I claim that, for any $k \ge 1$, $\gamma^k(G)^{n^k}\gamma^{k+1}(G)/\gamma^{k+1}(G) \le \gamma^k(K)\gamma^{k+1}(G)/\gamma^{k+1}(G)$. Since $G$ and hence $\gamma^c(G)$ are torsion free, that will prove that $\gamma^c(K)$ is nontrivial, and hence that $K$ has class $c$.

We prove the claim by induction on $k$. The case $k=1$ is clear from the definition of $K$, so assume the result is true for $k-1$ and we will prove it for $k$. Now $\gamma^k(G)/\gamma^{k+1}(G)$ is generated by elements $[x,y]$ with $x \in X$ and $y$ in a generating set of $\gamma^{k-1}(G)$, so $\gamma^k(G)^{n^k}\gamma^{k+1}(G)/\gamma^{k+1}(G)$ is generated by the images of $[x,y]^{n^k} = [x^n,y^{n^{k-1}}]$ (mod $\gamma^{k+1}(G)$).

But, by the inductive hypothesis, the elements $y^{n^{k-1}}$ (mod $\gamma^k(G)$) lie in $\gamma^{k-1}(K)$, and $x^n \in K$, so we get the claim for $k$.