By a result of Laczkovich ('[Analytic subgroups of the reals](http://dx.doi.org/10.1090/S0002-9939-98-04241-5)' Proc AMS Vol 126 (1998)), any non-open analytic subgroup of a Polish locally compact group can be covered by countably many closed Haar null sets. 


Is this result of Laczkovich true for any Polish (not necessarily locally compact) groups?
More precisely:


**Problem 1:** Does each non-open analytic subgroup $H$ of a Polish abelian group $G$ belong to the $\sigma$-ideal $\mathcal E$ generated by closed Haar null subsets of $G$?


A Borel subset $A$ of a Polish group $G$ is Haar null of there exists a Borel probability measure $\mu$ on $G$ such that $\mu(xAy)=0$ for all $x,y\in G$. 


A related weaker question also seems to be open.


**Problem 2:** Does each non-open analytic subgroup $H$ of a Polish abelian group $G$ belong to the $\sigma$-ideal $\mathcal M\_h$ generated by closed Haar-meager subsets of $G$?


A Borel subset $A$ of a topological group $G$ is called *Haar-meager* if there exists a continuous map $f:K\to G$ defined on a compact metrizable space $K$ such that $f^{-1}(xAy)$ is meager in $G$ for all $x,y\in G$. It can be shown that a closed subset $A$ of a topological group $G$ is Haar-meager if and only if there exists a compact subset $K\subset G$ such that for every $x,y\in G$ the intersection $K\cap xAy$ is nowhere dense in $K$. It is easy to see that each closed Haar-null set is Haar-meager. More information on Haar meager sets can be found in the paper "[On Haar meager sets](http://arxiv.org/pdf/1403.1909.pdf)" by Darji.



Can each non-open analytic subgroup of a Polish abelian group be covered by countably many closed Haar null subsets?
The answer to both problems (1 and 2) is negative: the Polish group $G=\mathbb Z^\omega$ contains a dense meager Borel subgroup $H$ (which can be written as the difference $H=A\setminus B$ of two $F\_\sigma$-sets in $G$) which cannot be covered by countably many closed Haar-meager subsets of $G$. Such subgroup $H$ is constructed in Example 6.3 of [this paper](http://arxiv.org/abs/1509.09073).


The subgroup $H$ also gives a (partial) counterexample to the question [Meager subgroups of compact groups](https://mathoverflow.net/questions/207533/meager-subgroups-of-compact-groups/215547#215547) .


Indeed, for any countable cover $\{A\_n\}\_{n\in\omega}$ of the group $H$ the closure $\bar A\_n$ of some set $A\_n$ is not Haar-meager (and hence not Haar-null) in $G$ and hence $\bar A\_n\bar A\_n^{-1}$ is a neighborhood of zero, which implies that $A\_nA\_n^{-1}$ is not nowhere dense in $H$.


