For a group $G$ generated by a finite set $S$ we denote by $B\_{G,S}(n)$ the ball of radius $n$, that is the set of all elements in $G$ which are expressible as products $x\_1x\_2\ldots x\_n$ where $x\_i\in S\cup S^{-1}\cup\{1\}$. One calls the set $Q$ generic in $G$ with respect to $S$ if $$\lim\_{n\to\infty}\sup \frac{|Q\cap B\_{G,S}(n)|}{|B\_{G,S}(n)|}=1.$$


My question is whether there exist a group $G$ and a proper subgroup $H<G$ such that $H$ is a generic subset in $G$ with respect to some finite generating set $S$ of the group $G$.



Generic set that is a proper subgroup
No, it's not possible. For notational convenience assume $S$ is symmetric and $1\in S$, so that $B\_{G,S}(n) = S^n$. Suppose
$$
 |H\cap S^{n\_i}|/|S^{n\_i}|\to 1
$$
for some subsequence $(n\_i)$. Let $x$ be an element of $S$ not in $H$. Since $H$ and $Hx$ are disjoint we have
$$
 |H\cap S^{n\_i}x^{-1}|/|S^{n\_i}| = |Hx\cap S^{n\_i}|/|S^{n\_i}| \to 0.
$$
Thus in particular
$$
 |S^{n\_i} \cap S^{n\_i} x^{-1}|/|S^{n\_i}| \to 0.
$$
But
$$
 S^{n\_i-1} x \subset S^{n\_i},
$$
so
$$
 S^{n\_i-1} \subset S^{n\_i} \cap S^{n\_i} x^{-1}.
$$
Also $S^{n\_i} \subset S^{n\_i-1} S$. Thus
$$
 |S^{n\_i} \cap S^{n\_i} x^{-1}| / |S^{n\_i}| \geq |S^{n\_i-1}|/|S^{n\_i}| \geq 1/|S| \not\to 0,
$$
a contradiction.


But I suppose you can have subgroups $H$ with $\limsup |H\cap S^n|/|S^n|$ arbitrarily close to $1$? However, we would need to have (1) lots of generators, (2) exponential growth. (If $|S^n|$ has subexponential growth then you have an invariant measure so you can't get bigger than upper density $1/2$.) Something in the free group $F\_n$?


