Finitely generated, periodic group such that each conjugacy class is finite must be finite?

Question

This is essentially a repeat of this question. However, the OP didn't seem to put in any work towards a solution and didn't provide context. Nevertheless, I'm still interested in the solution, and I am having difficulty.

To repeat:

Let $G$ be a finitely generated, periodic group such that each conjugacy class is finite. Does it follow that $G$ is finite?

Here's what I have done which was spurred by Derek Holt's hint in the previous question:

Let $G$ be generated by $x_1, \ldots, x_n$.

Since each conjugacy class is finite this means that $C_G(x)$ has finite index in $G$ for all $x\in G$. We also have in general that if two subgroups have finite index, then their intersection has finite index. Thus $C_G(x_1)\cap\cdots\cap C_G(x_n)$ has finite index in $G$.

This intersection is none other than $Z(G)$ since every element of $G$ can be written as a finite product of the $x_i$. So we know that $Z(G)$ has finite index.

I have yet to use the fact that each element of $G$ has finite order, but don't immediately see how that is going to help.

I am mostly interested in a hint to get me to the solution. Any help is appreciated.

Answer 1

I want to provide a quick answer to your weaker question (the one you actually post, as opposed to the one in the comments). I don't want to answer the "full" version, as I think some time should pass before re-asking and answering a closed question (say, 5 days).

So, you proved that $Z(G)=C_G(x_1)\cap\ldots C_G(x_n)$ has finite index in $G$. We'll start from there.

1. Finite-index subgroups of finitely generated groups are also finitely generated, so $Z(G)$ is finitely generated.
2. $Z(G)$ is abelian.

The above two facts, along with the assumption that $G$ is torsion, implies that $Z(G)$ is finite. As $Z(G)$ has finite index in $G$, $G$ must also be finite. This proves your result.