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Motivation: Let $G$ be any group such that all its subgroups are finitely generated, then it is easy to see that $G$ satisfies ACC.

We know there are finitely generated groups that don't satisfy the above, for example see this answer. So my question is can we have an example of a finitely generated group which does not satisfy ACC?

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If $G$ is a free group in two generators, there is a subgroup $F$ of $G$ which is isomorphic to a free group in countably many generators; for example, we could take $F=G'$ the derived subgroup of $G$.

Suppose the generators are $\{x_n:n\in\mathbb N\}$. Pick any bijection $\phi:\mathbb N\to\mathbb Q$, and for each real number $r\in\mathbb R$ let $H_r$ be the subgroup of $G$ generated by the set $\{x_n:\phi(n)<r\}$.This provides us with an uncountable set $\{H_r:r\in\mathbb R\}$ of proper subgroups of $G$ such that $$r<s\implies H_r\subsetneq H_s.$$ You can easily find now an ascending chain of subgroups of $G$ which does not stabilize!

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    $\begingroup$ You could consider much more simply the chain of subgroups $(K_n)_{n\geq1}$ with $K_n$ generated by $x_1$, $\dots$, $x_n$, of course; but the example above shows that the situation is much worse than having ACC fail :-) $\endgroup$ – Mariano Suárez-Álvarez May 1 '14 at 19:34

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