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Sep 14, 2019 at 16:17 comment added Sean Eberhard @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.
Sep 14, 2019 at 5:18 comment added Sean Eberhard @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.
Sep 14, 2019 at 0:48 comment added user6976 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?
Sep 13, 2019 at 14:50 comment added Sean Eberhard True, thanks. I actually had in mind $H_1 \cap H_2$ positive-density -- I've added that variant now too.
Sep 13, 2019 at 14:48 history edited Sean Eberhard CC BY-SA 4.0
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Sep 13, 2019 at 14:10 comment added YCor You want to assume $G$ infinite to avoid trivial counterexamples.
Sep 13, 2019 at 13:40 history asked Sean Eberhard CC BY-SA 4.0