Periodic center implies periodic in a nilpotent group Some background before I get into the problem - I've been out of school for a little while, but decided to head back to pursue a  doctorate in math. In the meantime, I've been going back through some old homework to try and brush up on some of the theory. For this particular course, I no longer have my lecture notes, and there are no doubt gaps in my knowledge that have grown over the years. From memory, we loosely followed Rotman's group theory text - as such my notation follows his (stuff operates on the left e.g. $x^y = yxy^{-1}$, $[x,y] = xx^y$, etc.).
Problem statement: Prove that if $G$ is nilpotent and $Z(G)$ is periodic then $G$ is periodic.
My solution idea:
Lemma 1: In any group $G$, if $Z(G)$ is periodic then $Z_2(G)$ is periodic.
Proof: Suppose $x \in G$, $y \in Z_2(G)$. Then $[x,y] \in Z(G)$. Let $n = |[x,y]|$. Then $[x,y]^n = 1 = [x,y^n]$. Hence $y^n \in Z(G)$. By hypothesis, there exists $m \in \mathbb{N}$ such that $(y^n)^m = 1$ so that $|y| \leq mn.$ $\square$
EDIT:: Lemma 2: If  $G$ is nilpotent of class $c+1$, then $\frac{G}{Z(G)} = \bar{G}$ is nilpotent of class $m \leq c.$
Proof: For each $i \in \mathbb{N}$, define $\bar{Z_i} = \frac{Z_{i+1}}{Z_1}$, where $Z_1 = Z(G).$ By the correspondence theorem, $\bar{Z_i} \triangleleft \bar{G}$. Moreover, $\frac{\bar{Z}_{i+1}}{\bar{Z_i}} \cong \frac{Z_{i+2}}{Z_{i+1}} = Z(\frac{G}{Z_{i+1}}) \cong Z(\frac{\bar{G}}{\bar{Z_i}}).$ Hence $\frac{\bar{Z}_{i+1}}{\bar{Z}_i} \leq \frac{\bar{G}}{\bar{Z}_i}.$ Therefore, $1 = \bar{Z}_0 \triangleleft \bar{Z}_1 \triangleleft  ... \triangleleft \bar{Z}_c = \bar{G}$ is a central series of $\frac{G}{Z_1}.$ Since $\bar{Z}_{c-1} \neq \bar{G}$, we deduce that this series is has length $c$. Thus $\bar{G}$ is nilpotent of class $m \leq c.$ $\square$
Returning to the main problem, we proceed by induction on the nilpotence class of $G$. If $G$ has class 1, then $Z(G) = G$ and the result holds. Suppose now that the result holds for all groups of class $c$. Then if $G$ has class $c+1$, we have shown that $\bar{G}$ has class $c$. Now $Z(\bar{G}) = \frac{Z_2}{Z_1}$. By Lemma 1, $Z_2$ is periodic, so that $Z(\bar{G})$ is periodic. By induction, $\bar{G}$ is periodic. Thus if $g \in G$, there exists $n \in \mathbb{N}$ such that $g^n \in Z(G).$ But then there exists $m \in \mathbb{N}$ such that $(g^n)^m = 1$ meaning $|g| \leq mn.$ Therefore, by induction, every nilpotent group is periodic. $\square$
EDIT: I updated Lemma 2 with a fix. Is my proof sound now?
 A: Here is my proposed counterexample:
$G = \langle x,y_k,z_k\,(k>0)\mid y_k^{2^k}=z_k^{2^k}=[x,y_k]=z_k, [x,z_k]=1,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
[y_j,y_k]=[y_j,z_k]=[z_j,z_k]= 1\, (\forall j,k>0)\,\rangle.$
Then $G$ is nilpotent of class 2, $Z(G) = \langle z_k : k > 0 \rangle$ is periodic, but $x$ has infinite order.
The result would be true for finitely generated nilpotent groups.
A: As Derek Holt points out, the result does not hold in general. The issue with your proof, as explained by Arturo Magidin in the comments, is that in Lemma 1 you actually proved the following:

If $y\in Z_2(G)$ then $\forall\:x\in G,\:\exists\:n\in\mathbb{Z}\:\text{s.t.}\:[x, y]^n=1$.

However, you need to swap the quantifiers, so need to prove the following:

If $y\in Z_2(G)$ then $\exists\:n\in\mathbb{Z}\:\text{s.t.}\:\forall\:x\in G,[x, y]^n=1$.

You can swap these quantifiers around if and only if $Z(G)$ has bounded torsion/period ($\exists n\in\mathbb{Z}\:\text{s.t.}\:\forall\: g\in Z(G), g^n=1$). Notice that Derek Holt's counter-example does not have bounded period. Therefore, (assuming I have not missed any other errors in what you wrote) what you have proven is the following, due to S. Dixmier*:

Theorem. If $G$ is nilpotent of class length $c$ and $Z(G)$ has bounded period $n$, then $G$ has bounded exponent dividing $n^c$.

Now, it is well-known that the periodic elements in a nilpotent group $G$ form a subgroup $t(G)$, and indeed that if $G$ is finitely generated then $t(G)$ is finite. We therefore have the following, surprisingly strong, corollary:

Corollary. If $G$ is finitely generated and $Z(G)$ is periodic then $G$ is finite.

Proof. As $G$ is finitely generated, $t(G)$ is finite. By the above theorem, $G=t(G)$ so $G$ is finite. QED

*S. Dixmier, Exposants des quotients des suites centrales descendants et ascendants d'un groupe, C.R. Acad. Sci. Paris 259 (1964), pp. 2751--2753. (I found this reference in R.B. Warfield, Jr. Nilpotent Groups, Springer-Verlag Lecture Notes in Mathematics 513. I can find no electronic copy of this paper though.)
A: This is probably close to Derek's example, but I'm not fond of the tradition of defining groups with presentations, so here's another approach.
Let $H_3(\mathbf{Q})$ be the (Heisenberg) group of matrices
$$\begin{pmatrix}1 & x & z \\ 0 & 1 & y\\ 0 & 0 & 1\end{pmatrix}:\quad x,y,z\in\mathbf{Q}$$
and $K$ its subgroup of elements with $x=y=0$ and $z\in\mathbf{Z}$. The proposed counterexample is $G=H_3(\mathbf{Q})/K$.
Indeed, let $Z$ be the center of $H_3(\mathbf{Q})$, namely those elements with $x=y=0$ (with $z$ arbitrary). So $Z/K\simeq\mathbf{Q}/\mathbf{Z}$ is torsion. To conclude, it is enough to check that the center of $G$ is reduced to $Z/K$. Indeed, if a matrix $\xi$ as above is not in $Z$, say $x\neq 0$ (the case $y\neq 0$ being analogous), take the commutator with the matrix $\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 1/2x\\ 0 & 0 & 1\end{pmatrix}$ to get the element $\begin{pmatrix}1 & 0 & 1/2 \\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}$, which is not in $K$, so $\xi$ is not central modulo $K$, and hence the center of $G$ is indeed reduced to the torsion group $Z/K$. (While clearly $G$ is not torsion.)
