Skip to main content
7 events
when toggle format what by license comment
May 24, 2022 at 7:53 comment added IJL @YCor: great idea to get it down to $F_4$.
May 24, 2022 at 0:02 vote accept tota
May 24, 2022 at 0:02
May 23, 2022 at 16:38 comment added tota Thanks@IJL for your answer.
May 23, 2022 at 16:27 comment added YCor And I guess the smallest $F_n$ we can thus get is indeed $F_4$, since every finite subgroup of $\mathrm{Aut}(F_n)$ for $n\le 3$ is solvable.
May 23, 2022 at 16:26 comment added YCor Nice. If instead of the free product $A_5\ast A_5$ you take the double $A_5\ast_{A_4}A_5$, you still have a canonical split surjection onto $A_5$, and this time, by the same argument, you get $F_4=F_{5-1}$ as kernel (as you get 5 free orbits instead of 60 free orbits of edges).
May 23, 2022 at 15:06 history edited IJL CC BY-SA 4.0
added 531 characters in body
May 23, 2022 at 14:58 history answered IJL CC BY-SA 4.0