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Does there exist a infinite group besides $\Bbb Z$ that all its nontrivial subgroups have finite index?

I know $\Bbb Z$ works, but is there any other examples? Or a proof that $\Bbb Z$ is the only one with such properties.

On one hand I can't gave a proof: the best I can do is to observe that every element have infinite order with the infinite cyclic group having finite cosets, therefore any other element must have a power going into the cyclic group, in other words every two element must have some kind of "least common multiple" which is similar to the integers. But I can't proof any commutative thing and was stuck.

On the other hand I doubt maybe there exist some weird free group that satisfy the properties, but also I can't gave a construction.

Q Lee
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    All non-trivial subgroups of $\mathbb Z$ have finite index, you mean? – David A. Craven Mar 27 '23 at 10:34
  • @DavidA.Craven of course subset Z has index n as an integer (btw I'm not familiar with mse, can I type latex here? I do not know how to type the \mathbb Z). – Q Lee Mar 27 '23 at 10:44
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    I mean you made a typo in your question. – David A. Craven Mar 27 '23 at 10:44
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    Yes ${\mathbb Z}$ is the only example. To prove this let $H$ be a maximal cyclic subgroup. Then $H$ must be self-centralizing, or else there would be an abelian subgroup with torsion elements, so $|G:H| \le 2$, but if $|G:H|=2$ then $G$ is the infinite dihedral group, which is not an example. So $G=H$. – Derek Holt Mar 27 '23 at 11:50
  • @DerekHolt why a group has to have a maximal cyclic subgroup? – freakish Mar 27 '23 at 12:06
  • @DerekHolt thank you very much, but I found problem filling in the gaps of this sketch, such as why the centralizer of H in G must be H itself, and why |G:H|≤2 is a consequence. May I ask for a further explanation? thx! – Q Lee Mar 27 '23 at 12:35
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    @freakish The subgroups all have finite index so just take one of smallest index. – Derek Holt Mar 27 '23 at 13:16
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    @QLee I already explained why $H$ must be self-centralizing. If $g \in C_G(H) \setminus H$, then $\langle H,g\rangle$ is abelian, but then maximality of $H$ means that it has nontrivial torsion subgroup. Then $|G:H| \le 2$ follows from the fact the automorphism group of an infinite cyclic group has order $2$. You still have not corrected the typo in your question. – Derek Holt Mar 27 '23 at 13:19
  • https://math.stackexchange.com/questions/379234/terminology-for-infinite-groups-all-of-whose-subgroup-have-finite-index/380075#380075 – Mikko Korhonen Mar 28 '23 at 00:25
  • @DerekHolt very grateful! but during the last step, what about the group $G=<a,b>$ with presentation $a^2=b^2$ and $aba=b$, it's not actually the infinite dihedral group, but satisfies $[G:]=2$ I think? Is this an example? – Q Lee Mar 28 '23 at 05:25
  • @MikkoKorhonen also thank you for the link shared! I have tried to search for similar questions but failed at first. Maybe I need more time to go over the proof posted there, my algebra is very bad. – Q Lee Mar 28 '23 at 05:36
  • @QLee That's a finite group (the dihedral group of order 8). ($bab^{-1}=a^{-1} \Rightarrow ba^2b^{-1}=a^{-2}$, but $a^2=b^2$, so $b^2=b^{-2} \Rightarrow b^4=1$.) And you still have not corrected the typo in the question. – Derek Holt Mar 28 '23 at 08:07

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This has been answered in the comments, but I will make it into a proper answer.

Let $G$ be an infinite group in which every nontrivial subgroup has finite index. We claim that $G$ is an infinite cyclic group, and hence is isomorphic to $({\mathbb Z},+)$.

Let $1 \ne x \in G$ and $H = \langle x \rangle$. Since $H$ has finite index in $G$ and $G$ is infinite, $H$ must be infinite. So $G$ is torsion-free. Note also that the core of $H$ in $G$ has finite index, and so we can assume that $H$ is a normal subgroup of $G$. Choose $H$ such that it is normal in $G$ and has minimal index subject to that condition.

Let $C = C_G(H)$. Then $H \le Z(C)$ and so $|C:Z(C)|$ is finite and hence, by a result of Schur, the derived group $[C,C]$ is finite. But then we must have $[C,C]=1$, so $C$ is abelian, and by the fundamental theorem of abelian groups $C$ is infinite cyclic, and then by the maximal choice of $H$ we have $C=H$.

Suppose that $H \ne G$, and let $y \in G \setminus H$. Then, since $y \not\in C_G(H)$, we must have $y^{-1}xy=x^{-1}$. Then $y^{-2}xy^2=x$, so $y^2 \in C_G(H) = H$, and hence $y^2 = x^k$ for some $k \in {\mathbb Z}$. So $y$ centralizes $x^k$, but $y^{-1}x^ky= x^{-k}$,so $x^{2k}=0$ and hence $k=0$ and $y^2=1$, contradiction, since $G$ is torsion-free.

So $G=H$ as claimed.

Derek Holt
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  • Fantastic! I'm sorry that it was beyond my ability to fill in the gaps of the sketch in the comment although you have tried your best to hint me. But really thank you for the effort to make it into an answer and this time I fully recognize the proof. – Q Lee Mar 28 '23 at 10:24
  • In the second-to-last paragraph, how do you know $y$ normalizes $H$? Your argument seems to show only $N_G(H) = H$. – Sean Eberhard Mar 28 '23 at 13:17
  • @SeanEberhard I have corrected the argument, but unfortunately I have had to use the result of Schur on subgroups having centre of finite index, which I think was also used in the earlier answer to this question that you found. I expect the question should be closed as a duplicate, but I wonder whetehr it can be answered without using Schur's Theorem. – Derek Holt Mar 28 '23 at 13:55
  • I also felt that Schur's theorem was probably overkill. (And to put credit where credit's due, the earlier question was found by @lulu, https://math.stackexchange.com/questions/4667279/infinite-group-with-all-subgroups-having-finite-index/4667942?noredirect=1#comment9864893_4667279.) – Sean Eberhard Mar 28 '23 at 14:10
  • (or rather Mikko, https://math.stackexchange.com/questions/4667279/infinite-group-with-all-subgroups-having-finite-index/4667942?noredirect=1#comment9864098_4667279) – Sean Eberhard Mar 28 '23 at 14:12
  • @SeanEberhard yes the proof there with Schur theorem seems to overcome the problem AT LAST... btw after commenting 3 hours ago for "fully recognizing" lol, I tried to write it down and was stuck in the point here you pointed out. But unfortunately I think it should finally be something trivial enough which I should do it myself instead of asking repeatedly, and the following hours was spent on it with obviously fruitless effort lol – Q Lee Mar 28 '23 at 14:31
  • also I discussed with my schoolmate and found some introduction of Schur's Theorem which may be useful for anyone in need, since I cannot found it directly on wiki. It seems that this question is much harder than I think, and maybe I should question my tutor for putting it in the homework lol. – Q Lee Mar 28 '23 at 14:45
  • I have the opposite point of view regarding your tutor: It sounds like you learned a lot of mathematics while thinking about this problem! – Lee Mosher Mar 28 '23 at 15:51