Timeline for An endomorphism of free groups
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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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 |