Timeline for Is there a 7-regular graph on 50 vertices with girth 5? What about 57-regular on 3250 vertices?
Current License: CC BY-SA 3.0
9 events
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| Jan 30, 2012 at 17:33 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 2082 characters in body
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| Jan 30, 2012 at 3:16 | comment | added | Noam D. Elkies | Possibly the most natural way to see the action of $S_5$ is to identify the $10$ vertices with $5 \choose 2$ pairs and say two pairs are adjacent iff they're disjoint. (In other words, the Petersen graph is the complement of the "triangle graph $T_5$".) | |
| Jan 29, 2012 at 23:28 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
added 268 characters in body
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| Oct 28, 2010 at 21:36 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
correction
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| Oct 28, 2010 at 21:32 | comment | added | Aaron Meyerowitz | Aha, thanks for the correction, I'll put it in. | |
| Oct 28, 2010 at 18:18 | comment | added | Chris Godsil | Aschbacher only proved the automorphism group could not be a rank-3 group, ie, the graph on 3250 vertices could not be distance transitive. G. Higman proved that it could not be vertex transitive. Martin Mačaj and Jozef Širáň recently showed that order of the group is at most 375. | |
| Oct 28, 2010 at 18:09 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
S6 not A6
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| Oct 28, 2010 at 11:51 | comment | added | Robert Bell | The automorphism group of the Petersen graph is isomorphic to the symmetric group on 5 elements. (There is a nice exercise in chapter one of Meier's book "Groups, Graphs, and Trees" which outlines a proof that the automorphisms freely permute the five 4-vertex induced subgraphs which are totally disconnected.) | |
| Oct 28, 2010 at 3:45 | history | answered | Aaron Meyerowitz | CC BY-SA 2.5 |