# Rank gradient in free products amalgamating a finite subgroup

Let $A,B$ be finitely generated groups with a common finite subgroup $C$. Suppose that $[A : C] > 2, [B : C] > 1$.

Must $A *_C B$ have positive rank gradient?

See Which 3-manifolds have positive rank gradient? for a definiton of rank gradient.

The assumption on the index is necessary (otherwise we take $A,B = \mathbb{Z}/2\mathbb{Z}, C = \{1\}$).

The case $C = \{1\}$ has been established by Marc Lackenby - Proposition 3.2 of http://people.maths.ox.ac.uk/lackenby/er190804.pdf

I think that there should be a proof using Bass-Serre theory, at least in the case that $A,B$ are residually finite. The argument is as follows:

Since $A,B$ are residually finite, $G = A *_C B$ is residually finite as well. It is enough to show that the rank of normal subgroups $N$ of finite index in $G$ grows linearly with the index $[G : N]$. Residual finiteness, allows us to take $N$ to be disjoint from $C$ in $G$. From Bass-Serre theory, one can hopefully write $G$ as a free product with amalgamation (since $N$ does not meet any conjugate of $C$ in $G$) and calculate its rank.

• You should make a residual finiteness assumption to avoid trivial counterexamples (and also assume explicitly the implicit $[B:C]>1$). Also please give a definition, or link to another post, for the definition of rank gradient. – YCor Nov 16 '15 at 16:44
• @YCor - I have referenced a definition. I do not see which counterexamples arise if either $A$ or $B$ (or both) are not residually finite. Furthermore, I think that it is sufficient to assume only that $[A : C] > 2$ and not also $[B : C] > 2$. In the absence of residual finiteness, the resulting free product with amalgamation may have no finite index subgroups at all, but in this case, the rank gradient is positive by definition. – Pablo Nov 16 '15 at 16:53
• If $A$ and $B$ both have no nontrivial finite quotient, then $A\ast B$ has no nontrivial finite quotient as well, so the rank gradient is zero. – YCor Nov 16 '15 at 16:56
• Actually I had no doubt about the fact that fg groups with infinitely many ends have positive first $\ell^2$ Betti number, but the fact that fg groups with positive first $\ell^2$ Betti number have positive rank gradient is not exactly what is stated in the reference I gave. I guess it's known anyway. – YCor Nov 17 '15 at 0:09
• @Pablo, a (fg) group with infinitely many ends is precisely a group which splits over a finite group and is not virtually cyclic. This is a famous theorem of Stallings. – HJRW Nov 17 '15 at 9:27