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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

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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.

  • $\begingroup$ Why does $|S|=3$? $\endgroup$ – gaoxinge Feb 20 '14 at 5:45
  • 6
    $\begingroup$ P is a $3$-regular and $Cay(G,S)$ is $|S|$-regular. $\endgroup$ – mehranian Feb 27 '14 at 11:16
  • $\begingroup$ 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? $\endgroup$ – sha Dec 2 '16 at 20:51
  • $\begingroup$ @mehranian Did you meant $D_5$ ? $\endgroup$ – sha Dec 5 '16 at 23:18

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