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I know that if $G$ is a finitely generated group, then $G$ has at most countably many finite index subgroups. Is this result still true if $G$ is countably generated?

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No. For example, let $G$ be the additive group of sequences $(a_n)$ of elements of $\mathbb{Z}/2\mathbb{Z}$ such that $a_n=0$ for all but finitely many $n$.

Then for each of the uncountably many non-zero sequences $(b_n)$ of elements of $\mathbb{Z}/2\mathbb{Z}$ (with possibly infinitely many non-zero $b_n$), there is a subgroup $$\left\{(a_n):\sum a_nb_n=0\right\}$$ of $G$ of index $2$.

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  • $\begingroup$ I didn't read the question carefully enough. Nice counter-example! $\endgroup$ – user1729 Jul 23 '14 at 15:15
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    $\begingroup$ By the way, another (arguably more natural but less elementary) counterexample is the free group on countably many generators, which has uncountably many homomorphisms to $\mathbb{Z}/2\mathbb{Z}$, and therefore uncountably many subgroups of index $2$. $\endgroup$ – Jeremy Rickard Jul 23 '14 at 17:39
  • $\begingroup$ @JeremyRickard : From the less elementary point of view, your "arguably" word is the correct one, since when one considers the abelian group case (i.e. $\mathbb Z$-modules), the first types of modules one can think about are the easiest i.e. the free ones. I do agree with your comment $\endgroup$ – Patrick Da Silva Jul 23 '14 at 23:23

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