Must a group of exponential growth is virtually free?

Question

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.

Answer 1

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.