Let $G$ be a finitely generated group. As we know that if $G$ contains a non-abelian free group of finite index, then $G$ has exponential growth. Does the converse of it holds? i.e. if $G$ is of exponential growth does it imply that $G$ must contain a non-abelian free group of finite index.
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$\begingroup$ I know that lamplighter group has exponential growth but I don't know whether it contains a non-abelian group or not. But I guess it does not. $\endgroup$– LeventDec 3, 2016 at 17:06
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$\begingroup$ @Levent it's solvable, so has no non-abelian free subgroup at all. $\endgroup$– YCorDec 5, 2016 at 1:12
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$\begingroup$ Sunny, a theorem of Stallings asserts that a torsion-free virtually free finitely generated group is actually free. Yet for many examples it can be shown very easily that some group is not virtually free. For instance, the direct product of a non-abelian free group and an infinite cyclic group has the property that every finite index subgroup contains a direct product of a non-abelian free group and an infinite cyclic group, and hence is not virtually free. $\endgroup$– YCorDec 5, 2016 at 1:17
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$\begingroup$ A group can be torsion and have exponential growth. Look for Burnside groups. $\endgroup$– Mustafa Gokhan BenliFeb 5, 2018 at 10:02
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$\begingroup$ Can you add link to refrerence? $\endgroup$– SunnyFeb 5, 2018 at 10:34
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1 Answer
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No. By a famous result of Gromov, a group has polynomial growth if and only if it has a nilpotent subgroup of finite index. There are groups of intermediate growth, i.e. between polynomial and exponential, but they are unusual. So any group that you can think of that is not virtually nilpotent is very likely to have exponential growth.
As examples you could take solvable groups that are not virtually nilpotent, such as the group $\langle x,y,z \mid yz=zx, y^x=z, z^x=yz \rangle$ or hyperbolic groups that are not virtually free such as $\langle x,y \mid x^l=y^m=(xy)^n=1 \rangle$ with $1/l+1/m+1/n<1$, or a surface group.
Added later: As another example, which clearly has exponential growth and is also clearly not virtually free, take a free product of a noncyclic free abelian group and a nonabelian free group.
answered Dec 3, 2016 at 17:08
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$\begingroup$ but what is the guarantee that they are not virtually free? $\endgroup$– SunnyDec 4, 2016 at 5:37
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$\begingroup$ @SunnyRathore: A solvable group cannot be virtually free unless it is virtually cyclic. Surface groups are never virtually free (assuming the surface has genus $>0$). Derek: You surely meant $1/l + 1/m + 1/n<1$, not $=1$. $\endgroup$ Dec 4, 2016 at 5:45
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