Let $p$ be prime. Consider the group $G$ given by the presentation
$$P=\langle a,b,\{ x_m\mid m\in\Bbb N\}\mid a^p, \{x_m^{p^m}, x_m=b^mab^{-m}\mid m\in\Bbb N\}\rangle.$$
My question is two-fold:
Are the only finite orders of elements in $G$ in the set $$X=\{p^k\mid k\in\Bbb N\cup\{0\}\}?$$
If so, how do I show it?
The reason this is a two-fold question is that I'm fairly sure the answer to (1) is "yes", so I'm more concerned with (2).
Thoughts:
It seems like we can rely on (generalising) this standard proof that for free groups, we have $F_r\le F_2$. The relations $x_m=b^mab^{-m}$ can be removed by Tietze transformations to give $G$ isomorphic to the group given by
$$Q=\langle a,b\mid a^p, \{(b^mab^{-m})^{p^m}\mid m\in\Bbb N\}\rangle.$$
Now I wave my hands a little and say something like, by looking at $P$ and $Q$, we get
$$G\stackrel{{\rm approx.}}{\cong}\left((\Bbb Z\ast\Bbb Z_p)\ast\left({\huge*}_{m\in\Bbb N}\Bbb Z_{p^m}\right)\right)/H,\tag{$*$}$$
where the $\Bbb Z$ comes from $b$, $\Bbb Z_p$ comes from $a^p$, $\Bbb Z_{p^m}$ comes from $x_m$, and $H$ is some leftover stuff from $x_m=b^mab^{-m}$; and it is my intuition, based on $P$ and $Q$, that no orders of finite elements exist for $G$ outside $X$. The free products suggest this to me.
The best I can do from here is say, "just look at $(*)$!"
Please help.
If $x_m$ is conjugate to $a$, then the order of $x_m$ coincides with the order of $a$, that is, the order of $x_m$ is $p$. It follows that the finite orders of elements of group $G$ are $p$ and $1$.