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Let $G$ be a finitely generated residually finite group with positive rank gradient, and let $F_2$ be the free group on $2$ elements. Must there be an embedding $i \colon F_2 \to G$ ?

A group $G$ is called residually finite if the intersection of all of its finite index subgroups is trivial.

A group $G$ is said to have positive rank gradient if the rank of finite index subgroups grows linearly with the index. More formally, the Rank Gradient (RG) of a finitely generated group $G$ is defined to be:

$$\mathrm{RG}(G) = \inf_{H} \frac{\mathrm{rank}(H) - 1}{[G : H]}$$ where $H$ runs over all subgroups of finite index in $G$, and the rank of a group is the smallest cardinality of a generating set for the group.

The notion of rank gradient is also important in the theory of $3$-manifolds.

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    $\begingroup$ Silly remark: "residually finite" (RF) is unnecessary, since if it's true in the RF case, letting $N$ be the intersection of finite index subgroups of $G$, we can apply the result to $G/N$ (noting that RG$(G)=$RG$(G/N)$) and lift $F_2$ to $G$. $\endgroup$ – YCor Sep 1 '15 at 13:44
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In

Denis Osin, Rank gradient and torsion groups. Bull. Lond. Math. Soc. 43 (2011), no. 1, 10–16,

the following theorem is proved

Theorem For every prime $p$, there exists a finitely generated infinite residually finite $p$-group with positive rank gradient.

http://arxiv.org/abs/0905.1322

http://blms.oxfordjournals.org/content/43/1/10

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  • $\begingroup$ I am not completely sure how to parse the statement of the theorem. Should "residually finite $p$-group" be read together, meaning that the intersection of subgroups of index a power of $p$ is trivial ? $\endgroup$ – Joël Sep 1 '15 at 14:37
  • $\begingroup$ @Joel But I think "(residually finite) $p$-group" is equivalent to "residually (finite $p$-group)". $\endgroup$ – Derek Holt Sep 1 '15 at 14:46
  • $\begingroup$ I am still confused. What is a $p$-group? For me it is a finite group of order a power of $p$. $\endgroup$ – Joël Sep 1 '15 at 14:56
  • $\begingroup$ p-group means each element has order a p-power. It need not be finite. Like the Grigorchuk group $\endgroup$ – Benjamin Steinberg Sep 1 '15 at 15:11
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    $\begingroup$ It is worth mentioning that Jan-Christoph Schlage-Puchta proved a similar result independently, see his paper: A p-group with positive rank gradient. J. Group Theory 15 (2012), no. 2, 261–270. You might also like to look at a joint paper with me: On p-deficiency in groups. J. Group Theory 16 (2013), no. 4, 497–517. (Both can be found on the Arxiv.) $\endgroup$ – Yiftach Barnea Sep 1 '15 at 18:33
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Lackenby ( http://people.maths.ox.ac.uk/lackenby/lg070105.ps ) proved that a finitely presented group, which has a pro-$p$ completion of positive rank gradient, is large, i.e. it contains a subgroup of finite index that projects onto a non-abelian free group. So while the answer to the original question is no, if you add some extra conditions it becomes yes in a very strong sense.

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  • $\begingroup$ I see that in the paper there are two more conditions (apart from positivity of the rank gradient) needed to conclude that the group is large. Is your claim correct? $\endgroup$ – Pablo Sep 6 '15 at 9:56
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    $\begingroup$ Sorry, when thinking about rank gradient I am usually only interested in the pro-$p$-case, so I remembered Lackenby's result n a pro-$p$ way. I corrected my answer accordingly. $\endgroup$ – Jan-Christoph Schlage-Puchta Sep 10 '15 at 20:04

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