Is there an infinite non-commutative group $G$ such that every proper subgroup of $G$ is commutative?
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asked Mar 28, 2018 at 14:59
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1$\begingroup$ 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. $\endgroup$– Jeremy RickardCommented Mar 28, 2018 at 15:08
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$\begingroup$ Thanks - I'll reformulate my question, or remove it altogether $\endgroup$– Dominic van der ZypenCommented Mar 28, 2018 at 15:10
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7$\begingroup$ 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). $\endgroup$– YCorCommented Mar 28, 2018 at 15:16
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$\begingroup$ Note that groups as in the question have the property that it's generated by any non-commuting pair. $\endgroup$– YCorCommented Mar 28, 2018 at 15:21
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2$\begingroup$ 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. $\endgroup$– Denis TCommented Mar 28, 2018 at 16:10
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