A group as the union of finitely generated subgroups

Question

Suppose a group $G$ satisfies

(1) $G=\bigcup_{i\ge 1} G_i$, where $G_i \le G_{i+1}$ and all are subgroups of $G$;

(2) there exists a number $m$ such that each $G_i$ is generated by at most $m$ elements.

Can we prove that $G$ is finitely generated?

Answer 1

This is false. Consider, for instance, $G = \mathbb Q$ and $G_i = \frac{1}{i!} \mathbb Z$. Each $G_{i!}$ is generated by $1/{i!}$, but $\mathbb Q = \bigcup \frac{1}{i!} \mathbb Z$. Furthermore, I claim that $\mathbb Q$ is not finitely generated. Indeed, take a finite subset $\emptyset \neq S \subseteq \mathbb Q$. Let $n$ be the least common multiple of the denominators of the elements of $S$. Then $\langle S\rangle \subseteq \frac{1}{n} \mathbb Z \neq \mathbb Q$, so $\mathbb Q$ is not finitely generated.