Left syndeticity and right syndeticity in nilpotent group

Question

$\DeclareMathOperator\Pf{\mathcal{P}_\mathrm{f}}$Question: Does there exist any reference regarding the study of left and right syndeticity in nilpotent group? More specifically, did anyone introduce/study before in any article whether left and right syndeticity are equal or not in nilpotent groups?

To elaborate the question here I recall the notion of left and right syndeticity [Definition 4.38] from the book 'Algebra in the Stone-Cech compactification' by Hindman and Strauss.

Definition: Let $S$ be a semigroup. Let $A\subseteq S$.

1. The set $A$ is right syndetic if there exists some $G\in\Pf(S)$ such that $S=\bigcup_{t\in G}t^{-1}A$.
2. The set $A$ is left syndetic if there exists some $G\in\Pf(S)$ such that $S=\bigcup_{t\in G}At^{-1}$.

Note that $\Pf(S)$ stands for the collection of all nonempty finite subsets of $S$.

Thanks in advance for any help or suggestion.

Answer 1

Let me call "left cobounded" and "right cobounded" subsets of a group which you call "right syndetic" and "left syndetic" respectively. That is, $X\subset G$ is left cobounded if there exists $F$ finite such that $G=FX$, etc.

I claim the following:

Proposition. Let $G$ be a countable group with at least one infinite conjugacy class. Then there is a subset of $G$ that is left cobounded but not right cobounded.

This applies in particular to every countable non-abelian, torsion-free nilpotent group.

Proof. Fix a proper length $|\cdot|$ on $G$. Fix an element $w$ with an infinite conjugacy class. Then there is a sequence $(x_n)$ such that $|x_n^{-1}wx_n|$ tends to infinity. Extracting if necessary, we can suppose that $|x_{n+1}|-|x_n|$ also tends to infinity. Choose $(r_n)$ tending to infinity, such that $r_n=o(|x_{n+1}|-|x_n|)$, $r_n=o(|x_{n}|-|x_{n-1}|)$ and $r_n=o(|x_n^{-1}wx_n|)$. Define $Y=\bigcup_n B_{\mathrm{right}}(x_n,r_n)$, where $B_{\mathrm{right}}(x,r)=xB(r)$ and $B(r)$ is the $r$-ball around $1$. Define $X=G\smallsetminus Y$. Clearly $X$ is not right cobounded (the right distance of $x_n$ to $X$ tends to infinity). I claim that $X$ is left cobounded. By contradiction, assume the contrary: $X$ is not left cobounded.

Choose $x_n\theta_n$ in $B_{\mathrm{right}}(x_n,r_n)$ that maximizes the left distance to $X$. Then for infinitely many $n$ (say $n\in J$), we have $wx_n\theta_n\notin X$. That is, for $n\in J$ we have $wx_n\theta_n\in Y$. Since $||wx_n\theta_n|-|x_n||\le r_n+|w|$ and the latter is $o(|x_{n+1}|-|x_n|)$ and $o(|x_{n}|-|x_{n-1}|)$ and $(|x_n|)$ is increasing, for large enough $n\in J$ we have $wx_n\theta_n\in B_{\mathrm{right}}(x_n,r_n)$, say $wx_n\theta_n=x_n\theta'_n$. Thus for large enough $n\in J$, we have $x_n^{-1}wx_n=\theta'_n\theta_n^{-1}$, and thus, $|x_n^{-1}wx_n|\le 2r_n$, which is a contradiction since $r_n=o(|x_n^{-1}wx_n|)$.

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Remark: the converse is true: if $G$ is a group with only finite conjugacy classes ("FC-group") and $X$ is left cobounded, then $X$ is also right cobounded. Indeed, write $G=FX$ with $F$ finite. Then $G=XF'$ with $F'=\bigcup_{g\in G}g^{-1}Fg$, and $F'$ is also finite.