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?


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 $2$.

  • $\begingroup$ I didn't read the question carefully enough. Nice counter-example! $\endgroup$ – user1729 Jul 23 '14 at 15:15
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.