Timeline for Are groups with every proper, non-trivial subgroup infinite cyclic simple?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Feb 28, 2022 at 17:38 | vote | accept | ADL | ||
| Feb 22, 2022 at 19:46 | answer | added | Giles Gardam | timeline score: 9 | |
| Jun 24, 2021 at 14:34 | comment | added | Mikko Korhonen | @AchimKrause: If $G/Z(G)$ is finite, then the commutator subgroup $[G,G]$ is finite, by a theorem of Schur. | |
| Jun 24, 2021 at 11:34 | comment | added | ADL | @AchimKrause I was meaning trying to adapt the general proof of the case when $G/\langle x\rangle$ is finite to when $G/\langle x\rangle$ is torsion. It is a standard fact that if $G$ is torsion-free, $\langle x\rangle$ infinite cyclic and $G/\langle x\rangle$ is finite and then $G$ must be infinite cyclic - so can the proof be adapted? | |
| Jun 24, 2021 at 11:19 | comment | added | Achim Krause | @YCor: Cool, I wasn't aware of Tarski monsters. So there are plenty of groups that could occur as $G/\langle x \rangle$. I suppose we are precisely looking for a Tarski monster $H$ and a cohomology class in $H^2(H;\mathbb{Z})$ whose image in $H^2(C_p;\mathbb{Z})$ is nontrivial for all subgroups $C_p$. | |
| Jun 24, 2021 at 11:14 | comment | added | Achim Krause | @ADL I think we can rule out the case of $G/\langle x\rangle$ finite: As discussed above, it must be a simple group (since the center $\langle x\rangle$ is the maximal normal subgroup of $G$), and every proper subgroup is cyclic. It cannot itself be cyclic, as then $G$ would be abelian. If $y\in G/\langle x\rangle$ has order $2$, it is either central or we find a nontrivial conjugate $y'$, and then $y, y'$ generate a dihedral subgroup. So $G/\langle x\rangle$ is either dihedral or odd, in both cases it cannot be simple. | |
| Jun 24, 2021 at 11:01 | comment | added | YCor | I believe one also might try to indeed construct such examples of central extensions of Tarski monsters. | |
| Jun 24, 2021 at 10:58 | comment | added | ADL | @AchimKrause Yes, I was getting tangled up with groups of this form too. I tried to mimic the proof that if $G$ is torsion-free and $G/\langle x\rangle$ is finite then $G$ is cyclic, and the idea here is to prove that $G$ splits. So for example if $H^2(G/\langle x\rangle, \mathbb{Z})=0$ then we'd be done, but I cannot see why this would be $0$. | |
| Jun 24, 2021 at 10:39 | comment | added | Achim Krause | Also note that if we do have nontrivial center $\langle x \rangle$, then $G/\langle x \rangle$ is a pretty weird group: It has the property that every proper subgroup is finite cyclic. It feels like one should be able to finish from here, but I haven't figured out how yet. | |
| Jun 24, 2021 at 10:06 | comment | added | Achim Krause | Small observation: Any normal subgroup is necessarily central. Assume $\langle x \rangle$ is normal, and there is $y$ with $yxy^{-1} = x^{-1}$. Then $\langle x,y\rangle$ cannot be cyclic, so it must be all of $G$. The subgroup $\langle x,y^2\rangle$ is abelian, thus a proper subgroup, and must be cyclic. So $G$ is a normal extension $\mathbb{Z}\to G \to C_2$ with sign action on $\mathbb{Z}$, and thus isomorphic to the infinite dihedral group, which contains $2$-torsion, contradiction. | |
| Jun 24, 2021 at 9:49 | history | edited | ADL | CC BY-SA 4.0 |
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| Jun 24, 2021 at 9:43 | history | edited | ADL | CC BY-SA 4.0 |
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| Jun 24, 2021 at 9:34 | history | edited | ADL | CC BY-SA 4.0 |
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| Jun 24, 2021 at 9:33 | comment | added | ADL | @MikaeldelaSalle I meant to exclude that possibility! I'll edit the question to rule it out. | |
| Jun 24, 2021 at 9:32 | comment | added | Mikael de la Salle | What about an infinite cyclic group? | |
| Jun 24, 2021 at 9:28 | history | edited | ADL | CC BY-SA 4.0 |
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| Jun 24, 2021 at 9:22 | history | asked | ADL | CC BY-SA 4.0 |