Index of center $Z(G)$ is finite implies the number of elements of conjugacy class is finite

Exercise

Let $G$ be a group such that its center $Z(G)$ has finite index in $G$. Show that every conjugation class has finite elements.

I don't know how to attack the problem. I thought the following: if $C$ is a conjugacy class, then $C$ can be thought as an orbit under the action of $G$ on itself by conjugation. Could it be that the orbit divides $[G:Z(G)]$?, I am not so sure if this is true, if this was the case then from here it follows the statement of the problem. I would appreciate hints or ideas to solve the problem.

• Yes you are right, the size of the orbit divides $|G:Z(G)|$. – Derek Holt Sep 18 '14 at 21:31

Hint. The size of a conjugacy class $x^G$ is equal to the index $[G:C_G(x)]$, where $C_G(x)$ is the centraliser of $x$ in $G$.
• Let me see: it is easy to see that $Z(G)≤C_G(x)$. So $|Z(G)|$ divides $|C_G(x)|$. How could I show from here that $[G:C_G(x)]$ divides $[G:Z(G)]$? – user16924 Sep 18 '14 at 22:57
• It's not necessary (though true). Since $Z(G)\leq C_G(x)$, and since $[G:Z(G)]$ is finite by assumption, therefore, $[G:C_G(x)]$ is also finite. On the other hand, I don't see anything that allows you to say that $\left|Z(G)\right|$ or $\left|C_G(x)\right|$ is finite. (But, it doesn't matter.) – James Sep 19 '14 at 1:13