Motivation and Details:
Thinking back a couple of years ago, when I did combinatorial group theory a lot, I decided to explore the idea that whenever an element of a group has finite order $n$, then there exist elements in that group of orders each power of $n$.
This lead me to the following definition of mine:
Definition: Let $G$ be a group. If for all $n\in\Bbb N$, if $g\in G$ with $|g|=n$ we have for all $m\in\Bbb N$ there is at least one $h_m\in G$ with $|h_m|=n^m$, call $G$ powerful-element.
These groups exist:
Example: Consider $G=\Bbb Z$ or $$G=\prod_{m=1}^\infty \Bbb Z_{p^m}$$ for prime $p$.
The Question:
Let $G$ be a powerful-element group that is not torsionfree${}^\dagger$. Can $G$ be finitely presented?
Thoughts:
Of course, $G$ is infinite.
I thought for a while that I had found a finitely generated, powerful element group that is not torsion free. Here is the candidate:
Let $p$ be prime. Consider the group $G$ given by the presentation
[. . .]
$$Q=\langle a,b\mid a^p, \{(b^mab^{-m})^{p^m}\mid m\in\Bbb N\}\rangle.$$
But that didn't work: see this question of mine.
So I'm stuck.
My guess is that, no, no such group exists, but I wouldn't be surprised if one did; so I'm really hedging my bets on this one.
Context:
This is the sort of problem I think I could answer myself, provided luck and a few months' time. Like I said: I did a lot of combinatorial group theory during a PhD attempt (that turned into an MPhil for a number of reasons); but now I'm doing a postgraduate research degree in linear algebraic groups and I don't have enough time.
The kind of answer I'm looking for is an complete solution. This could be asking too much, though, given the nature of how some problems in the area are easy to state but exceptionally difficult to solve.
Please help :)
$\dagger$: That is, there does exist a torsion element: an $x\in G$ of finite order.
