Following section 7.1 in Peterson-Thom's paper here, we say a countable group $G$ is boundedly generated by the subgroups $G_1, \cdots, G_n$, if there exists an integer $k\in\mathbb{N}$, such that every element in $G$ is a product of less that $k$ elements from $G_1,\cdots, G_n$.
Note that this is NOT the commonly used definition, where $G_i$'s are assumed to be cyclic.
My question is the following.
If G is finitely generated and has infinitely many ends, does it hold that G is not boundedly generated by $G_1, \cdots, G_n$ for any subgroups $G_i$'s with infinite index (or may be relaxed to just any proper subgroups)?
Remark: This question is motivated by this one asked here.
The "conjecture" is false, for example take $F_2= \langle x,y \rangle$, then $F_2$ is boundedly generated with respect to $\langle x \rangle, \langle y \rangle,$ and $[F_2,F_2]$, since every element in $F_2$ is equal to $x^ny^m c$ for some $c \in [F_2,F_2]$ and some $n,m \in \mathbb{Z}$, since $x^ny^m$ form a complete set of coset representatives i.e. $F_2 = \bigcup_{n,m \in \mathbb{Z}} x^ny^m[F_2,F_2]$. (You could come up with many examples using the same idea, where $F_2/N$ is boundedly generated then take $N$ and representatives)
There are restrictions on what sort of groups you can be boundedly generated with respect to. Note that groups which have infinetly many ends are acylindrically hyperbolic. By work of Osin (Prop 1.7), at least one of the groups, $G_1,...,G_k$, needs to be acylindrically hyperbolic. For example, a group which has infinitely many ends can not be boundedly generated with respect to a finite collection of amenable groups.