Timeline for Exponent of the quotient of the commutator of a free group
Current License: CC BY-SA 3.0
4 events
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| Mar 9, 2016 at 6:08 | history | edited | user114539 | CC BY-SA 3.0 |
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| Mar 8, 2016 at 10:15 | comment | added | Derek Holt | The answer is $2$ for $p=2$ and $3$ for $p=3$. It would probably be possible to compute it for $p=4$ and $6$ (but was $p$ supposed to be prime?). I presume the question is mainly concerned with other values of $p$, when $F/F^p$ is either known to be infinite or at least not known to be finite. | |
| Mar 8, 2016 at 10:03 | comment | added | Alireza Abdollahi | If $p=2$, then $[F,F]/[F^2,F]$ is generated by the image of $[x,y]$ in the quotient group. So the quotient is cyclic if $p=2$. Why you think that the exponent must be finite in the case $p=2$? $x$ and $y$ are free generators of $F$. | |
| Mar 8, 2016 at 5:42 | history | asked | user114539 | CC BY-SA 3.0 |