If every nonzero quotient of $G$ is infinite, then $G$ is divisible

Question

I tried proving this by contradiction (assuming $G$ is abelian, else $G$ could not possibly be divisible):
If $G$ is not divisible then there exist $g\in G$ and $n\in \mathbb {N}$ such that for all $y\in G$, $y^n\ne g$. My next goal would have been to show that, since $S=\{y^n :y\in G\}\ne G$, then $G/S$ is a nonzero quotient yet finite, to complete the proof by the contrapositive. However, all I managed to prove was that $G/S$ has exponent $n$. If I assumed that $G/S$ is finitely generated then I could use the classification theorem for abelian groups and argue that $G/S$ is the direct product of some finite number of cyclic groups of order at most $n$, which would imply $G/S$ is finite.
If $G/S$ is not finitely generated then I have no idea how else to proceed. A direct proof is equally hard since I have little experience with infinite groups. Maybe I am missing something, any hints or ideas are appreciated.

Answer 1

Actually, there are nonabelian divisible groups, even finitely-generated ones (V. S. Guba, Finitely generated complete groups, Math. USSR-Izv. 29 (1987), no. 2, 233–277.). For an easier infinite example, consider $SO(n)$, $n\ge 3$: It is an infinite simple divisible group, see this question.

Anyway, even if you assume that your group is abelian, the statement you are trying to prove is false. Consider, for instance, the additive group of dyadic rational numbers, i.e. numbers of the form $$ \frac{a}{2^b}, \quad a\in {\mathbb Z}, \quad b\in {\mathbb N}. $$ These groups are not divisible, say, by 3 (I hope, that's clear), but every nontrivial quotient is infinite (a nice exercise).