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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 $G$.

Al Tal
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1 Answers1

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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 $F_n$?

Sean Eberhard
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  • 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. – Al Tal Apr 06 '16 at 10:41
  • 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 $|H\cap S^{n_i}|/|S^{n_i}| \to 1$. – Sean Eberhard Apr 06 '16 at 10:50
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    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. – Ilya Bogdanov Apr 06 '16 at 12:35
  • @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 $1/[G:H]$. – Sean Eberhard Apr 06 '16 at 12:53
  • Is it possible to get something bigger than $1/2$ with the usual limit? (As in Ilya's example there is no convergence.) – Al Tal Apr 06 '16 at 12:58
  • @AlTal Exactly, I think not, because the Cesaro averages converge to $1/[G:H]$. I can supply details (I think) if you like. – Sean Eberhard Apr 06 '16 at 13:02
  • @Sean Eberhard: Sure! The details are very much welcomed. Could you please write them at this page? – Al Tal Apr 06 '16 at 15:15
  • On second thought I'm not sure whether that's true. I had in mind a proof that if $\mu^{n}$ is the distribution of the random walk after $n$ steps then $\mu^{n}(H)$ converges to $1/[G:H]$. This follows from looking at the random walk as a Markov chain on $G/H$. However for just the set $S^n$ I'm not sure, would be curious to know. – Sean Eberhard Apr 06 '16 at 16:21
  • @AlTal In fact I was wrong. There are groups of infinite index with well-defined density arbitrarily close to $1$. See http://mathoverflow.net/questions/243686/finite-index-iff-positive-density/244444#244444. – Sean Eberhard Jul 16 '16 at 13:52