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Let $G$ be an infinite finitely generated group and $S$ a generating set. Define density with respect to the sequence of balls $S^n$.

If $H_1, H_2 \leq G$ have positive density, must $H_1 \cap H_2$ be (a) nontrivial? (b) positive-density?

Obviously if $H_1$ and $H_2$ have finite index then so does $H_1\cap H_2$, so such subgroups are not relevant. It is possible for an infinite-index subgroup to have positive density, e.g., $F_2 \leq F_2 \times \mathbf{Z}$ (with the "summed" generating set). The group $G = F_2 \times F_2$ is a near miss: the two factors don't have positive density in $S^n$, but they do have density about $1/n$, which is fairly large.

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    $\begingroup$ You want to assume $G$ infinite to avoid trivial counterexamples. $\endgroup$
    – YCor
    Commented Sep 13, 2019 at 14:10
  • $\begingroup$ True, thanks. I actually had in mind $H_1 \cap H_2$ positive-density -- I've added that variant now too. $\endgroup$
    – Sean Eberhard
    Commented Sep 13, 2019 at 14:50
  • $\begingroup$ Do you know an example of a subgroup of infinite index and positive density in a) a group of exponential growth b) a group of superlinear growth c) any f.g. group? $\endgroup$
    – user6976
    Commented Sep 14, 2019 at 0:48
  • $\begingroup$ @MarkSapir If the balls satisfy $|S^{n+1}| / |S^n| \to 1$ then they define a Folner sequence, which ensures that each subgroup has density = 1/index. There are also some groups of exponential growth, such as free groups, where every subgroup of infinite index has zero density (see the references here: mathoverflow.net/questions/243686/…), but I don't have a very good understanding why. $\endgroup$
    – Sean Eberhard
    Commented Sep 14, 2019 at 5:18
  • $\begingroup$ @MarkSapir Sorry Mark I did not understand your comment, because as I already mentioned in the question the subgroup $F_2$ of $F_2 \times \mathbf{Z}$ has infinite index and positive density. $\endgroup$
    – Sean Eberhard
    Commented Sep 14, 2019 at 16:17

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