Petersen graph is not a Cayley graph

Question

How can I show that the Petersen graph is not a Cayley graph?

I don't know very much about Cayley graphs, I know that they are vertex-transitive, but so is the Petersen graph. It probably has to do with the group structure of $\Gamma$ in $Cay(\Gamma,S)$ (which is a group of order $10$, i.e., it is either the cyclic group $C_{10}$ or the dihedral group $D_{10}$). Then I am guessing we go for some sort of contradiction(?). My group theory is a bit rusty, I would very much appreciate someone showing me how to do this one!

score 10 · Answer 1 · answered Oct 25 '13 at 16:19

10

Let $ P\simeq Cay(G,S)$, then $|S|=3$ and $|G|=10$, then $G\simeq \mathbb{C_{10}}$ or $\mathbb{D_{10}}$, in both cases $G$ has cycle order 4, whereas $P$ has cycle of or 5.

answered Oct 25 '13 at 16:19

mehranian

207

Why does $|S|=3$? – gaoxinge Feb 20 '14 at 05:45
6

P is a $3$-regular and $Cay(G,S)$ is $|S|$-regular. – mehranian Feb 27 '14 at 11:16
Hi @mehranian Is there a direct way to show that there is a cycle of order 4? given one generator is idempotent and two more generators and 10 elements over all? – sha Dec 02 '16 at 20:51
@mehranian Did you meant $D_5$ ? – sha Dec 05 '16 at 23:18

Petersen graph is not a Cayley graph

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