# Petersen graph is not a Cayley graph [duplicate]

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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!

## marked as duplicate by quid♦Jul 17 '18 at 18:26

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.
• Why does $|S|=3$? – gaoxinge Feb 20 '14 at 5:45
• P is a $3$-regular and $Cay(G,S)$ is $|S|$-regular. – mehranian Feb 27 '14 at 11:16
• @mehranian Did you meant $D_5$ ? – sha Dec 5 '16 at 23:18