Timeline for Graphs $G$ with $G \cong \text{Aut}(G)$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 22 at 4:37 | history | edited | Dominic van der Zypen | CC BY-SA 4.0 |
bijective graph homomorphism -> graph isomorphism (subtle difference)
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| May 22 at 4:35 | comment | added | Dominic van der Zypen | Thanks @bof for your observation - I will correct this! | |
| May 21 at 17:10 | history | edited | Martin Sleziak |
edited tags
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| May 21 at 15:17 | answer | added | Zerox | timeline score: 12 | |
| May 21 at 14:18 | comment | added | Peter LeFanu Lumsdaine | One simple observation: for any $\newcommand{\Aut}{\mathrm{Aut}}G$, $\Aut(G)$ is always vertex-transitive, since for any $f, g$, left-multiplication by $gf^{-1}$ is an automorphism sending $f$ to $g$. So if $G \simeq \Aut(G)$, then $G$ must be vertex-transitive, for starters. | |
| May 21 at 13:39 | comment | added | Dominic van der Zypen | @Zerox Right, thanks, I also realized it and deleted my comment. So "it looks like" (for what it's worth) that there are no connected examples of connected graphs $G$ with more than $2$ points and $G\cong\text{Aut}(G)$ | |
| May 21 at 13:37 | comment | added | Zerox | @DominicvanderZypen $K_3$ is not, $\text{Aut} (K_3)$ has $2$ components, both isomorphic to $K_3$. | |
| May 21 at 13:18 | comment | added | Zerox | @JoelDavidHamkins The complete graph $K_2$ is a (trivial) connected example. | |
| May 21 at 13:12 | comment | added | Joel David Hamkins | He wants a connected graph. But in light of the paucity of examples, any $G$ with $G\cong \text{Aut}(G)$ is interesting. | |
| May 21 at 13:11 | comment | added | HenrikRüping | and the graph with two vertices and no edge. | |
| May 21 at 13:00 | comment | added | HenrikRüping | I only see the graph with one vertex. | |
| May 21 at 12:57 | comment | added | Joel David Hamkins | Ah, darn. You are right. So do we have any examples? | |
| May 21 at 12:55 | comment | added | HenrikRüping | There is also $x\mapsto -x$. Thus $Aut(\mathbb{Z})$ should have two components. | |
| May 21 at 12:43 | comment | added | Joel David Hamkins | It seems that the adjacency graph on the integers is a countably infinite instance, but I don't yet see how to make uncountable instances. | |
| May 21 at 11:49 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |