Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the following question

Is the dodecahedral graph $D$ a Cayley graph?

I would like to show that it is not and I am lazy hence I am looking for the most cheap way.

I see two approaches that could help me solve the this problem

  1. Compute $\rm{Aut}(D)$ and show that it has no regular subgroup of order 20. I can do that with a computer quickly but I don't see a quick way to accomplish it without making use of a computer.
  2. Use Sabidussi's Theorem. We know that if $D$ is a Cayley graph then $D \cong \rm{Cay}(G,S)$ for a group $G$ of order $20$ and an inverse closed subset $S \subset G$ of order 3. Furthermore we know that in this case $G$ cannot be abelian since otherwise $\rm{Cay}(G,S)$ has a $4$-cycle. There are three non-abelian groups of order $20$ and this step would now require to verify that for all three groups $G$ and respective values of $S$, $\rm{Cay}(G,S)$ is not isomorphic to $D.$ Again quite an annoying step.

So my question is: Is there any quick way to show that $D$ is not a Cayley graph or a slick way to finish one of the listed approaches?

share|cite|improve this question
You use the fact that $\mathrm{Aut}(D)$ is isomorphic to $\mathrm{Alt}(5)\times\mathbb{Z}_2$. It's not hard to show by hand that this has no regular subgroup. In general there is no easy way to decide if a graph is a Cayley graph. – Chris Godsil Jan 27 '13 at 19:55
Mr. Godsil, your book on algebraic graph theory is great! – H_B Nov 30 '15 at 6:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.