Timeline for An approximate version of $g^2 = e$ for all $g \in G$, implies $G$ is Abelian
Current License: CC BY-SA 3.0
5 events
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| Aug 25, 2015 at 16:25 | comment | added | Igor Rivin | @NeilStrickland Ah, indeed (though in many matrix groups you need either one or very few commutators to express any element in the commutator subgroup) | |
| Aug 25, 2015 at 16:09 | comment | added | Neil Strickland | That's true, but the commutator subgroup is the subgroup generated by the commutators, which is not just the set of commutators. | |
| Aug 25, 2015 at 16:03 | comment | added | Igor Rivin | @NeilStrickland We probably understand the OP differently. My understanding of his hypothesis is that every square is close to the identity, and thus (by the three squares argument) every commutator is close to the identity. | |
| Aug 25, 2015 at 15:59 | comment | added | Neil Strickland | This argument seems to show only that the commutator subgroup is generated by a small neighbourhood of the identity, not that it is contained in a small neighbourhood of the identity. | |
| Aug 23, 2015 at 17:18 | history | answered | Igor Rivin | CC BY-SA 3.0 |