Timeline for Automorphism group of a special commuting graph
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 23, 2017 at 8:49 | comment | added | YCor | @maryam you're right (except that your comment is weirdly placed here) I've edited to complete my answer | |
| Sep 15, 2017 at 12:55 | comment | added | maryam | @Ycor, thanks for the answer. I have confused. I think you have found automorphism group of a graph containing conjugacy class (12) not (12)(34)? | |
| Sep 12, 2017 at 6:00 | comment | added | Peter Heinig | For the record: the comment containing "For inexperience" I misspoke when saying "only on the integer partitions of $6$": the right thing to say is just "partitions of $6$". What I said is wrong on more than one account: the usual technical term for "ordered partition" is composition (see e.g. Stanley's book on enumerative combinatorics); second, "the ordered partition" sounds like there were only one such partition; third, most seriously, saying "depends on the ordered partitions" sounds like as if, say, the number of conjugacy classes depended on how to order the partition. | |
| Sep 11, 2017 at 7:50 | comment | added | Peter Heinig | @FriederLadisch: many thanks for the explanations. For the record: I misread Chris Godsil's comment. I read his comment "we know the outer automorphism group of $S_6$ acts on the graph", which indeed sums up the gist of the issue (though, pedantically speaking, one should say 'the outer automorphism group of $S_6$ acts faithfully on the vertex set of the graph') for the triviality 'we know that $S_6$ acts on the vertex set of the graph'. The latter (again: my misreading) does not contain enough information to justifiably say 'knowing the order is enough'. | |
| Sep 10, 2017 at 16:18 | comment | added | Frieder Ladisch | Another proof that $X$ is $\operatorname{Aut}(S_6)$-invariant: The elements in $X$ have centralizer of order $16$, the other involutions centralizer of order $48$. | |
| Sep 10, 2017 at 16:14 | comment | added | Frieder Ladisch | @PeterHeinig: In fact, $\operatorname{Aut}(S_6)$ acts on the graph, because the double transpositions are invariant under the automorphism group. To see this, note that the alternating group is invariant under $\operatorname{Aut}(S_6)$, as $A_6$ is the only normal subgroup of index $2$ in $S_6$, and the only elements of order $2$ in $A_6$ are just the double transpositions. | |
| Sep 10, 2017 at 13:49 | comment | added | Peter Heinig | [...] So then we know that $\mathrm{Aut}(\Gamma)$ is a finite group of order $1440$ containing $S_6$ as subgroup. Do we know anything more? Of course, by definition of 'automorphism group of a graph', we also known that $\mathrm{Aut}(\Gamma)$ is a permutation group on 45 letters. (Since the graph in question has 45 vertices.) So we know $S_6\leq\mathrm{Aut}(\Gamma)\leq S_{45}$. But do we know anything more at this point? There seem to o be 5958 candidates still. I fear I am missing something obvious, yet I do not see why 'knowing the order is enough'. Could you please briefly explain why? | |
| Sep 10, 2017 at 13:45 | comment | added | Peter Heinig | @ChrisGodsil: many thanks for pointing this out. The 'order taken together with context is enough to deduce the isomorphism type'-aspect was lost on me; it is interesting. I do not quite understand it though. I follow for a while, but not till the conclusion. Of course, from the definition of the OP's $\Gamma$ is is evident that $S_6\hookrightarrow\mathrm{Aut}(\Gamma)$, so, as you say, the question remains whether there are more automorphisms. And of course, if we believe Sage's output, then seeing $1440=2\cdot 6! = 2\cdot\lvert S_6\rvert$ tells us that there are more automorphisms. [...] | |
| Sep 10, 2017 at 12:33 | comment | added | Chris Godsil | In this case knowing the order is enough - we know the outer automorphism group of $S_6$ acts on the graph, the only question is whether there is more. Second, I think anyone who wants to find the automorphism group of such a graph should automatically turn to a computer, and I suspect this may be part of F.C.'s motivation for the title of their answer. | |
| Sep 10, 2017 at 9:48 | comment | added | Peter Heinig | So while this is certainly a nice and useful answer, one should remain aware of the following: (0) Sage's utterances are not mathematical proofs, (1) YCor's answer seems to be a correct mathematical proof and accomplishes quite a bit more than Sage here: YCor seems to have found the correct isomorphism type in the haystack of 5958 disctinct isomorphism types, (2) the initial 'Easy enough' is slightly grating (to me). I can understand how this happens (not being a native speaker), yet it seems to make this look easier than it actually is. | |
| Sep 10, 2017 at 9:44 | comment | added | Peter Heinig | For inexperience users: "conjugacy_class(...)" makes Sage create the set of permutations in G which are conjugate to, e.g., the representative $(12)(34)$. Hopefully needless to say, this does not depend on the representative, only on the ordered partition of $6$, which is why the argument $[2,2,1,1]$, which corresponds to $2+2+1+1$, is used here. Moreover, the OP asked for the structure of this automorphism group, not only its order. I take the GP wiki's word for it that there are a whopping 5958 isomorphism types of them. | |
| Sep 10, 2017 at 7:42 | history | answered | F. C. | CC BY-SA 3.0 |