# Do positive-density subgroups intersect nontrivially?

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.

• You want to assume $G$ infinite to avoid trivial counterexamples. – YCor Sep 13 '19 at 14:10
• True, thanks. I actually had in mind $H_1 \cap H_2$ positive-density -- I've added that variant now too. – Sean Eberhard Sep 13 '19 at 14:50
• 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? – Mark Sapir Sep 14 '19 at 0:48
• @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. – Sean Eberhard Sep 14 '19 at 5:18
• @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. – Sean Eberhard Sep 14 '19 at 16:17