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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 $G$.

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  • $\begingroup$ you want intersection in the numerator $\endgroup$
    – Benjamin Steinberg
    Apr 6, 2016 at 9:27

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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 $F_n$?

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  • $\begingroup$ Thanks! Nice argument. As I understand, it works for $\lim\sup$ (as in my question) too, even though you put it with the usual limit. $\endgroup$
    – Al Tal
    Apr 6, 2016 at 10:41
  • $\begingroup$ The limsup is covered by the subsequence $(n_i)$: $\limsup |H\cap S^{n}|/|S^{n}| = 1$ iff there is a subsequence $(n_i)$ such that $|H\cap S^{n_i}|/|S^{n_i}| \to 1$. $\endgroup$
    – Sean Eberhard
    Apr 6, 2016 at 10:50
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    $\begingroup$ On the additional question: surely yes. Let $G$ be the free group with $N$ generators, and let $H$ to be the subgroup of all irreducible words of even length. $\endgroup$
    – Ilya Bogdanov
    Apr 6, 2016 at 12:35
  • $\begingroup$ @IlyaBogdanov Nice example. On the other hand I think if you look at the Cesaro averages $\frac1n \sum_{k=0}^{n-1} |H\cap S^k|/|S^k|$, then in general these converge to $1/[G:H]$. $\endgroup$
    – Sean Eberhard
    Apr 6, 2016 at 12:53
  • $\begingroup$ Is it possible to get something bigger than $1/2$ with the usual limit? (As in Ilya's example there is no convergence.) $\endgroup$
    – Al Tal
    Apr 6, 2016 at 12:58

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