Timeline for Countable non-commutative groups such that all proper subgroups are commutative
Current License: CC BY-SA 3.0
9 events
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| Mar 28, 2018 at 20:28 | comment | added | YCor | @DenisT I don't know what you call a good approximation of "a good third" but there are about 5000 question tagged gr.group-theory and you'd have difficulty to find even 25 questions (half a percent) whose answer is given by Tarski monsters (and for which Tarski monsters are not part of the question). | |
| Mar 28, 2018 at 17:57 | comment | added | Dominic van der Zypen | Thanks for your comment @YCor. Maybe you can post this answer so we can close the question? Many thanks | |
| Mar 28, 2018 at 16:10 | comment | added | Denis T | It's pretty funny that a good third (maybe I'm exaggerating a bit) of group theory questions on MO have "Tarski monster" as answer. | |
| Mar 28, 2018 at 15:21 | comment | added | YCor | Note that groups as in the question have the property that it's generated by any non-commuting pair. | |
| Mar 28, 2018 at 15:16 | comment | added | YCor | Yes, Tarski monsters. These are nonabelian groups in which every proper subgroup is cyclic (finite or infinite). They were constructed by Olshanskii (Gromov says they are easy to construct in his hyperbolic book, but that it's much much harder when the constraint is to have cyclic subgroups of bounded order). | |
| Mar 28, 2018 at 15:11 | history | edited | Dominic van der Zypen | CC BY-SA 3.0 |
deleted 48 characters in body; edited title
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| Mar 28, 2018 at 15:10 | comment | added | Dominic van der Zypen | Thanks - I'll reformulate my question, or remove it altogether | |
| Mar 28, 2018 at 15:08 | comment | added | Jeremy Rickard | If $G$ is non-abelian and $x,y\in G$ with $xy\neq yx$ then the subgroup generated by $x$ and $y$ is countable and non-abelian. | |
| Mar 28, 2018 at 14:59 | history | asked | Dominic van der Zypen | CC BY-SA 3.0 |