Timeline for Let $S$ be the nonempty set of strongly regular graphs with given parameters. Must $S$ contain vertex transitive graph?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 25, 2016 at 7:55 | vote | accept | joro | ||
| Mar 24, 2016 at 14:09 | answer | added | Krystal Guo | timeline score: 9 | |
| Mar 24, 2016 at 12:05 | comment | added | Chris Godsil | A strongly regular graph on $p$ vertices ($p$ prime) is necessarily a conference graph, consequently $p\equiv1$ mod 4 and therefore there is a Paley graph on $p$ vertices. | |
| Mar 24, 2016 at 10:58 | comment | added | YCor | For many primes (e.g., $p=19$ by Jordan, see www-groups.mcs.st-and.ac.uk/~colva/dartmouth.pdf p.9), the only transitive groups on $p$ elements are symmetric, alternating or of affine type. So picking $v$ to be such a number, if you have a strongly regular connected graph on $v$ vertices that is not a complete graph, and whose automorphism group does not have a normal subgroup of order $p$, then you're done. | |
| Mar 24, 2016 at 10:34 | comment | added | joro | @YCor 1) Finite graphs. 2) All parameters, including the number of vertices: $(v,k,\lambda,\mu)$. | |
| Mar 24, 2016 at 10:30 | comment | added | YCor | 1) Is your question about finite graphs? 2) Does "given parameters" means that, in the language of en.wikipedia.org/wiki/Strongly_regular_graph, the integers $k$ (degree), $\lambda$ and $\mu$ (number of common vertices between any two adjacent, resp. non-adjacent vertices), are fixed? Or do you also fix the number $v$ of vertices [this is not a local assumption, so it sounds less natural]? | |
| Mar 24, 2016 at 9:30 | history | asked | joro | CC BY-SA 3.0 |