This is Exercise 2.6 of Roman's, "Fundamentals of Group Theory: An Advanced Approach". Searches in Approach0 were unsuccessful due to too many mathematical terms and a search on MSE for "Roman 2.6" returned nothing.
The Details:
Definition: A group $G$ is periodic (a.k.a., torsion) if each of its elements has finite order.
Definition: The centre $Z(G)$ of a group $G$ is defined by $$Z(G):=\{z\in G\mid zx=xz \;\forall x\in G\}.$$
The Question:
Let $Z<G$ be the centre of $G$. Show that if every element of $G$ not in $Z$ has finite order, then $G$ is periodic.
Thoughts:
This looks like the question is framed in a helpful way. Let $z\in Z$. It suffices to show that the order of $z$ is finite.
My intuition (and the previous exercise ibid.) suggests that the following lemma might help.
Lemma: For all $r,s$ in a group, $\lvert rs\rvert=\lvert sr\rvert$.
(The proof is routine.)
Another approach I'm thinking of is to take some $a\in G\setminus Z$ and "pit it against" our chosen $z$ in such a way that it forces the order of $z$ to be finite. To this end:
$$az=za\iff z=aza^{-1},$$
which doesn't tell us much; I was hoping that I could use the fact that conjugation by $a$ is an inner automorphism. But that's all I have there.
Yet another approach would be to exploit the following.
Lemma 2: $$Z(G)\unlhd G.$$
(The proof here is also routine.)
Lemma 3: $$G/Z(G)\cong{\rm Inn}(G),$$ where ${\rm Inn}(G)$ is the group of inner automorphisms under composition.
(For a proof, see Theorem 9.4 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)", p194.)
My hope is that whether $G/Z(G)$ is periodic or not might shed some light on whether $Z$ in question is periodic.
However, this machinery is not covered in the Roman's book so far and, given that the exercise occurs early on in the set of exercises for the chapter, there's likely to be a better, easier way. In fact, even quotient groups aren't covered yet.
Please help :)
Note: The first paragraph was written when it wasn’t entirely clear whether $Z<G$ meant proper subgroup or just subgroup.
The result as stated is false (unless we interpret $\lt$ to mean proper subgroup). If $G$ is abelian, then $G=Z(G)$, and the condition is vacuously satisfied; yet of course this does not tell you anything about whether $G$ is periodic or not. Taking a free abelian group (or just an abelian group with an element of infinite order) disproves the statement.
On the other hand, if we assume that $G$ is not abelian, then the result is fairly easy. As you note, it suffices to show that every central element has finite order. Let $z\in Z(G)$; since $G$ is not abelian, let $g\notin Z(G)$. Then $gz\notin Z(G)$ (since $Z(G)$ is a subgroup), and hence has finite order $n\gt 0$. Since $g\notin Z(G)$, it also has finite order $m\gt 0$. But because $z$ and $g$ commute, $$e = e^m = ((gz)^n)^m = (gz)^{nm} = g^{mn}z^{mn} = (g^m)^nz^{mn}=e^nz^{mn} = z^{mn}$$ so the order of $z$ is finite.