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Let $T=\mathrm{Tor}(G)$ be the torsion subset of $G$, i.e. $$T=\mathrm{Tor}(G):=\{g\in G: g^n=1 \text{ for some } n\geq 1\}.$$ In general $T$ is not a subgroup of $G$, so it doesn't make sense to talk about the "index" of $T$ in $G$. But we can still ask if finitely many translates can cover $G$.

A set $S\subseteq G$ is called syndetic if finitely many translates of $S$ can cover $G$, i.e. $$G=g_1S\cup\cdots \cup g_n S$$ for some $g_1,\ldots,g_n\in G$. If $S$ is a subgroup then this is equivalent to $[G:S]<\infty$.

Of course if $G$ is a torsion group then $G=T$ so of course $T$ is syndetic. Are there are any other ways that $G$ can be a finite union of translates of $T$?

Problem. Let $T:=\mathrm{Tor}(G)$ be the torsion subset of $G$. Suppose that $T$ is syndetic in $G$. Prove that $G$ is a torsion group.

This checks out when $G$ is finitely-generated abelian, because then $T$ is a direct summand of $G$ with $G=\mathbb{Z}^r\oplus T$. So the only way $[G:T]<\infty$ is if $r=0$ and $G=T$.

What about for nonabelian groups? If $G$ is finitely-generated and the commutator subgroup $G'$ has infinite index, we can bootstrap the f.g. abelian case to solve the problem.


Attempt

Maybe we should try to show that $T$ is a subgroup of $G$. Then $[G:T]<\infty$ which immediately implies that $G$ is a torsion group. It is also clear that the subgroup $\langle T\rangle$ generated by $T$ must have (literal) finite index.

An easy application of the Pigeonhole Principle shows that if $T$ has finite index in $G$, then: for every $g\in G$, there is some $k\in \mathbb{Z}$, $|k|\leq n$ such that $g^k=st$ is a product of two torsion elements $s,t\in T$.

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  • $\begingroup$ I don’t think the title truly reflects the question; yes, you can have the torsion subset have “finite index” (i.e., be syndetic), but your real question wasn’t whether this can happen in an infinite group, but rather whether it can happen in an infinite group that is not torsion... $\endgroup$ – Arturo Magidin Jan 3 at 0:00
  • $\begingroup$ P.S. I’ve only ever encountered “syndetic” in the context of subset of the natural numbers. Do you have a reference for this use? I’d be interested in having it. $\endgroup$ – Arturo Magidin Jan 3 at 0:01
  • $\begingroup$ You could look up Hindman--Strauss, "Density in Arbitrary Semigroups", or Bergelson--McCutcheon--Zhang, "A Roth Theorem for Amenable Groups". $\endgroup$ – Ehsaan Jan 3 at 2:35
  • $\begingroup$ Actually, a very good reference is Bergelson, "Minimal Idempotents and Ergodic Ramsey Theory". $\endgroup$ – Ehsaan Jan 3 at 4:43
  • $\begingroup$ Thank you for the references. $\endgroup$ – Arturo Magidin Jan 3 at 12:11
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Take the infinite dihedral group $D_{\infty} = \{r,s\mid s^2=1, sr=r^{-1}s\}$. The elements of $D_{\infty}$ are precisely the elements of the form $r^is^j$, with $i$ arbitrary and $j=0$ or $1$. The elements of finite order are precisely the identity, and those of the form $r^is$, which have order $2$. Thus, $T=\{r^is\mid i\in\mathbb{Z}\} \cup \{e\}$.

Since $r^i = s(r^{-i}s)\in sT$, then we have that $D_{\infty}=T\cup sT$. But $D_{\infty}$ is not a torsion group, since $r\in D_{\infty}$ has infinite order.

$D_{\infty}$ can also be realized as the semidirect product $\mathbb{Z}\rtimes\mathbb{Z}_2$, where the action is by inversion.

Note that this is an easy example that shows that the set of torsion elements need not be a subgroup in a nonabelian group (though it generates a characteristic subgroup since as a set it is invariant under automorphisms).

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  • $\begingroup$ I needed $D_{\infty} = \{r,s\mid s^2=1, srs^{-1}=r^{-1}\}$ and $r^i = sr^{-i} s^{-1}$ to convince myself $\endgroup$ – reuns Jan 3 at 0:18
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    $\begingroup$ @reuns: Convince yourself of what? I’m confused... $\endgroup$ – Arturo Magidin Jan 3 at 1:07

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