Suppose $G$ is any group and $X$ is the set of all elements of order $p$ in that group where $p$ is a prime. Prove that if $X$ is finite, then $\langle X\rangle$ is finite, where $\langle X\rangle$ is group generated by the elements of $X$.
If possible, I'm interested in a solution that assumes as few theorems as possible (aside from the well known ones of course).
On
Another way to prove this is by applying a classical result called Dietzmann's Lemma (for a proof see for example the book of Derek Robinson, Finiteness Conditions and Generalized Soluble Groups, Part 1.) It implies that if you have a finite set $X$ in a group $G$, only containing elements of finite order and is closed under conjugation, then $\langle X \rangle$ is finite.
You may as well assume that $G = \langle X \rangle$. Since the elements of $X$ have only finitely many conjugates, their centralizers of in $G$ have finite index in $G$, and hence so does their intersection, which is $Z(G)$.
So $|G:Z(G)|$ is finite. There is a theorem (which you would probably prefer to avoid using but for the moment I don't see how) that $|G:Z(G)|$ finite implies $G'$ is finite and then, since the result is clearly true for abelian groups, $G/G'$ is finite and hence so is $G$.