Let $\mathrm{cov}\_H(C\_2^\omega)$ be the smallest cardinality of a cover of the Boolean group $C\_2^\omega=(\mathbb Z/2\mathbb Z)^\omega$ by closed subgroups of infinite index. It can be shown that
$$\max\{\mathrm{cov}(\mathcal M),\mathrm{cov}(\mathcal N)\}\le \mathrm{cov}(\mathcal E)\le\mathrm{cov}\_H(C\_2^\omega)\le \mathfrak r,$$
where $\mathcal E$ is the $\sigma$-ideal generated by closed subsets of Haar measure zero in $C\_2^\omega$ (the upper bound $\mathrm{cov}\_H(C\_2^\omega)\le\mathfrak r$ is proved, for example, [here](https://mathoverflow.net/q/378850/61536)). It is known that $\max\{\mathrm{cov}(\mathcal N),\mathrm{cov}(\mathcal M)\}$ can be strictly smaller than $\mathfrak b$ (this happens, for example, in the Laver and Mathias models).



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> **Problem.** Is $\mathrm{cov}\_H(C\_2^\omega)<\mathfrak b$ consistent? What is the value of $\mathrm{cov}\_H(C\_2^\omega)$ in the Laver (or Mathias) model?
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Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgroups?
Lyubomyr Zdomskyy proved that in the Laver model $\mathrm{cov}\_H(2^\omega)=\omega\_1<\mathfrak b=\mathfrak c$.


His argument used the following known *Laver property* of the Laver model $V'$: for every
function $f:\omega\to\omega$ in $V'$ upper bounded by some function $h:\omega\to\omega$ in the ground model $V$, there exists $H:\omega\to [\omega]^{<\omega}$ in $V$ such that $|H(n)|\leq n+1$
and $f(n)\in H(n)$ for all $n\in\omega$.


Choose any increasing sequence $(k\_n)\_{n\in\omega}\in \omega^\omega\cap V$ such that $k\_0=0$ and $k\_{n+1}>k\_n+n+1$ for every $n\in\omega$. Consider the family $\mathcal H$ of closed nowhere
dense subgroups of $2^\omega$ of the form $ \prod\_{n\in\omega}H\_n, $ where
$H\_n$ is a proper subgroup of $2^{k\_{n+1}\setminus k\_n}$ and $(
H\_n)\_{n\in\omega}\in V$. We claim that for every $x\in 2^\omega$ there
exists $H\in\mathcal H$ with $x\in H$. Indeed, since $x{\restriction}
(k\_{n+1}\setminus k\_n)\in 2^{k\_{n+1}\setminus k\_n}$, the latter set
has size $2^{k\_{n+1}-k\_n}$, and the sequence $\langle
2^{k\_{n+1}-k\_n}\rangle\_{n\in\omega}$ lies in $V$, there exists a sequence
$\langle X\_n\rangle\_{n\in\omega}\in V$ such that
$$x{\restriction}(k\_{n+1}\setminus k\_n)\in
X\_n\in \big[ 2^{k\_{n+1}\setminus k\_n}\big]^{n+1}$$ for all $n\in\omega$. It
suffices to denote by $H\_n$ the subgroup of $2^{k\_{n+1}\setminus
k\_n}$ generated by $X\_n$ and note that $H\_n\neq 2^{k\_{n+1}\setminus
k\_n}$ because $|X\_n|=n+1$ and $k\_{n+1}-k\_n>n+1$.


Therefore in the model $V'$ we have $\mathrm{cov}\_H(2^\omega)\le|\mathcal H|\le|V\cap 2^\omega|=\omega\_1$.


