Finite-index iff positive density?

Question

Let $G$ be a finitely generated group and $S$ a symmetric generating set. Define density (lower density, say) with respect to the sequence of balls $S^n$.

Is it true that a subgroup of $G$ has positive density iff it has finite index?

Low-hanging fruit:

1. This is certainly true if $G$ has subexponential growth, as then $(S^n)$ is a Folner sequence.

2. In general, finite index implies positive lower density: Suppose $G = \bigcup_{i=1}^m g_i H$ where $m = [G:H]$. Then for every $n$ there is some $i$ such that $|g_iH\cap S^n| \geq |S^n|/m$. But $|g_i H\cap S^n| = |H \cap g_i^{-1} S^n| \leq |H \cap S^{n+r}|$, where $r$ is chosen so that $g_1,\dots,g_m \in S^r$. Thus $|H\cap S^{n+r}| \geq |S^n|/m \geq |S^{n+r}|/(m|S^r|)$, so $H$ has lower density at least $1/(m|S^r|)$.

There are a couple of results in the special case in which $G$ is a free group and $S$ is the standard set of generators. Woess [1] proves this when $H$ is normal in $F_r$, while [2, Theorem 2.5] extends this to non-normal $H$ (at the level of "private communication"). Judging from the speciality of these results one would guess that the answer is no in general, i.e., that at least in some groups with respect to some generating sets there are infinite-index subgroups of positive density. Is that true?

Lamplighter groups $\mathbf{Z}/2\mathbf{Z} \wr \mathbf{Z}^k$ are an interesting source of examples, but not quite counterexamples as far as I can tell.

[1] Wolfgang Woess, MR 731608 Cogrowth of groups and simple random walks, Arch. Math. (Basel) 41 (1983), no. 4, 363--370.

[2] Alexandre V. Borovik, Alexei G. Myasnikov, and Vladimir N. Remeslennikov, MR 2028100 Multiplicative measures on free groups, Internat. J. Algebra Comput. 13 (2003), no. 6, 705--731.

Answer 1

It's not true. Let $\def\Z{\mathbf{Z}} G = F_2 \times \Z$, let $H = F_2$, and let $S$ be the generating set consisting of the identity, $x,x^{-1},y,y^{-1}\in F_2$ and $1,-1\in\Z$. Then for all $n\geq 0$ and $k\in\Z$ we have $$ |S^n\cap(F_2 + k)| = \begin{cases} 2\cdot 3^{n-|k|} - 1 & \text{if}~ n\geq |k|, \\0 & \text{if}~n < |k|.\end{cases}$$ The reason is that the standard ball in $F_2$ of radius $n$ has size $2\cdot 3^n - 1$ (as there are $4\cdot 3^{m-1}$ words of length $m$ for each $m\geq 1$, and one of length $0$), but it first takes at least $|k|$ steps to get to $F_2 + k$ in the first place. Thus $|S^n| = 4\cdot 3^n - 2n - 3$ and $|S^n\cap F_2| = 2 \cdot 3^n - 1$, so $F_2$ has lower density $1/2$, but infinite index.

Moreover if you look at $G = F_r \times \Z$ and do the same calculation you find that $F_r$ has lower density $1-1/r$, so it is possible to have lower density arbitrarily close to $1$ but still have infinite index. On the other hand upper density equal to $1$ forces $H=G$: see Generic set that is a proper subgroup.