Recall that a group $G$ is boundedly generated if it can be written as a finite product of cyclic subgroups. And there are a lot of examples of groups that are (not) boundedly generated.
I am wondering whether it is true that if $G$ has infinitely many ends, then it is not boundedly generated.
Thanks to Paul's comments that brings Osin's work to my attention, now, I realized that I should have known the answer is yes from the following argument.
First, from Proposition 7.1 in Peterson-Thom's paper here, we know that if $G$ is boundedly generated in the above sense, then the first $\ell^2$-Betti number of $G$ is equal to zero.
Now, suppose $G$ has infinitely many ends, we know $G$ contains non-abelian free group $F_2$ as a subgroup (by Osin's paper above), hence $G$ is not amenable; then $H^1(G, \ell^2(G))=0$ from Peterson-Thom's above paper, so the subgroup $H^1(G, \mathbb{Z}G)=0$, which is equivalently to saying that $G$ has one-end, a contradiction.