Skip to main content

Timeline for An endomorphism of free groups

Current License: CC BY-SA 3.0

9 events
when toggle format what by license comment
Sep 26, 2017 at 20:48 answer added Andrew Clifford timeline score: 1
Sep 25, 2017 at 13:26 vote accept Stevie H
Sep 25, 2017 at 8:41 comment added Aurel @DerekHolt Indeed, I messed up my actions, thanks. What I meant was that $(1,2)(1,2,3,4)$ sends $1\mapsto 1$ and $2\mapsto 3$ and therefore conjugates $(1,2)$ to $(1,3)$ for a suitable normalisation of conjugation.
Sep 25, 2017 at 8:04 comment added Derek Holt @aurel $g_1=\alpha_1\alpha_2$ gives the same result as $g_1=\alpha_2$. I think you mean $g_1= \alpha_2\alpha_1$. Also, if you are suing right actions, then $(1,2)^{(1,2)(1,2,3,4)} = (1,2)^{(1,2,3,4)}=(2,3)$.
Sep 25, 2017 at 8:00 answer added YCor timeline score: 16
Sep 25, 2017 at 7:24 comment added Aurel @Seirios It is not clear to me. In fact, $S_4$ is generated by $(1,2)$ and $(1,2,3,4)$ but not by $(1,3) = (1,2)^{(1,2)(1,2,3,4)}$ and $(1,2,3,4)$, so in $F_2 = \langle\alpha_1,\alpha_2\rangle$, the endomorphism $\psi$ given by $g_1 = \alpha_1\alpha_2$ and $g_2 = 1$ is not an automorphism.
Sep 25, 2017 at 5:37 review Close votes
Sep 25, 2017 at 11:32
Sep 25, 2017 at 4:56 review First posts
Sep 25, 2017 at 5:50
Sep 25, 2017 at 4:54 history asked Stevie H CC BY-SA 3.0