$\newcommand\Sn{\mathit{Sn}}$A subset $A$ of a group $X$ is called *algebraic* if $A=\{x\in X: a\_0xa\_1x\dotsm xa\_n=1\}$ for some elements $a\_0,a\_1,\dotsc,a\_n\in X$.


Let $\mathcal A\_X$ be the family of all algebraic sets in $X$.


**Definition.** The *Steinhaus number* $\Sn(X)$ of an infinite group $X$ is the largest cardinal $\kappa$ such that for any cover $\mathcal C\subset\mathcal A\_X$ of $X$ with cardinality $\lvert\mathcal C\rvert<\kappa$ there exists a set $C\in \mathcal C$ such that $FCC^{-1}F=X$ for some finite set $F\subseteq X$.


By the famous Steinhaus–Weil Theorem, for any closed set $F$ of positive Haar measure in a compact topological group $X$ the set $FF^{-1}$ is a neighborhood of the identity in $X$.


This theorem implies that each compact topological group has $$\operatorname{cov}(\overline{\mathcal N}\_X)\le \Sn(X)\le \lvert X\rvert$$
where $\operatorname{cov}(\overline{\mathcal N}\_X)$ is the smallest cardinality of a cover of $X$ by closed sets of Haar measure zero.


Under Martin's Axiom, $\Sn(X)=\mathfrak c$ for each infinite compact Polish group.



> 
> **Problem.** Is $\Sn(X)=\mathfrak c$ for any infinite compact Polish group in ZFC?
> 
> 
> 


**Remark.** The answer is affirmative for commutative groups. This follows from the observation that each algebraic set in a commutative group is a coset of a subgroup $\{x\in X: x^n=1\}$.



Steinhaus number of a group
The answer to this problem is negative: *For the compact Polish group $X=S\_3^\omega$ we have $Sn(X)\le\mathfrak r$ where $$\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq [\omega]^\omega\;\wedge\;\forall f\in 2^\omega\;\;\exists R\in\mathcal R\;\;|f[R]|=1\}.$$*
By induction it can be shown $\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq[\omega]^\omega\;\wedge\;\forall f\in n^\omega\;\exists R\in\mathcal R\;\;|f[R]|=1\}$ for every $n\ge 2$. So, we can find a family $\mathcal R\subseteq[\omega]^\omega$ of cardinality $|\mathcal R|=\mathfrak r$ such that for every function $f:\omega\to S\_3$ there exists $R\in\mathcal R$ such that $|f[R]|=1$.
Then $$S\_3^\omega=\bigcup\_{R\in\mathcal R}\bigcup\_{g\in S\_3}A\_{R,g},\quad\mbox{where $A\_{R,g}=\{f\in S\_3^\omega:\{g\}=f[R]\}$}.$$
Observe that for every $R\in\mathcal R$ and $g\in S\_3$ the set $A\_{R,g}A\_{R,g}^{-1}=A\_{R,e}$ is nowhere dense in $S\_3^\omega$ and hence $FA\_{R,g}A\_{R,g}^{-1}F\ne S\_3^\omega$ for any finite set $F\subseteq S\_3^\omega$.


It remains to show that for every $R\in\mathcal R$ and $g\in S\_3$, the set $A\_{R,g}$ is algebraic in $S\_3^\omega$.


Let $d$ and $t$ be elements of order $2$ and $3$ in the symmetric group $S\_3$. It is easy to check that $S\_3=\{x\in S\_3:dxxdxx=e\}$ where $e$ is the identity of the group $S\_3$. On the other hand, $\{x\in S\_3:xtxxtx=t^2\}=\{e\}$ and hence $\{x\in S\_3:g^{-1}xtg^{-1}xg^{-1}xtg^{-1}x=t^2\}=\{g\}$.


Then $A\_{R,g}=\{x\in S\_3^\omega:a\_0xa\_1xa\_2xa\_3x=b\}$ where
$$a\_0(i)=a\_2(i)=\begin{cases}g^{-1}&\mbox{if $i\in R$};\\
d&\mbox{if $i\in\omega\setminus R$},
\end{cases}\quad
a\_1(i)=a\_3(i)=\begin{cases}tg^{-1}&\mbox{if $i\in R$};\\
e&\mbox{if $i\in\omega\setminus R$},
\end{cases}
$$
and
$$
b(i)=\begin{cases}t^2&\mbox{if $i\in R$};\\
e&\mbox{if $i\in\omega\setminus R$},
\end{cases}
$$


