Timeline for are there finite nonabelian characteristic quotients $G$ of $F_2$ inducing a surjection $Aut(F_2)\twoheadrightarrow Aut(G)$?

9 events

when toggle format	what		by	license	comment
Sep 29, 2016 at 13:22	vote	accept	stupid_question_bot
S Sep 27, 2016 at 22:52	history	suggested	Luc Guyot	CC BY-SA 3.0	Well, it's mostly about my name being right!
Sep 27, 2016 at 22:41	review	Suggested edits
S Sep 27, 2016 at 22:52
Sep 27, 2016 at 22:34	history	edited	Derek Holt	CC BY-SA 3.0	added 266 characters in body
Sep 27, 2016 at 22:07	comment	added	Arturo Magidin		Oh, okay; that explains it. The identity $[x^2,y]=1$ follows from $[x,y,z]=1$ and $[y,z]^2=1$, since $[x^2,y] = [x,y]^x[x,y]=[x,y][x,y,x][x,y]$.
Sep 27, 2016 at 20:28	comment	added	Derek Holt		Yes it has exponent $4$. The exponent-$p$-class (I should probably have hyphenated exponent-$2$) of a $p$-group is the length of its so-called $p$-central series, which is computed by the well-known $p$-quotient algorithm, and descends in elementary abelian layers. I think you also want the identity $[x^2,y]=1$.
Sep 27, 2016 at 20:02	comment	added	Arturo Magidin		Doesn't $F/K$ have exponent $4$ (though the images of the two generators have exponent $2$), as the subvariety of $\mathfrak{N}_2$ ought to correspond to the identities $x^4=[y,z]^2=[x,y,z] = 1$ ? (since a group of exponent 2 is necessarily abelian, after all).
Sep 27, 2016 at 19:32	history	edited	Derek Holt	CC BY-SA 3.0	added 240 characters in body
Sep 27, 2016 at 19:26	history	answered	Derek Holt	CC BY-SA 3.0