MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

From the literature, we know that the line graph of a complete graph $L(K_{q})$ is a Cayley graph if and only if $q \equiv 3$( mod 4) is a prime power. Now, if $q \equiv 3$( mod 4) is a prime power, then is it possible to construct a Cayley graph $Cay(G,S)$ with connection set $S=S^{-1}$ which is isomorphic to $L(K_{q})$. If so, How can we construct it with an explicit structure?

share|cite|improve this question
Thank you every one who answer this question – Mojtaba Jazaeri Apr 27 '14 at 11:42
See Chris Godsil's answer to… – Tony Huynh Apr 27 '14 at 13:55
Thanks a lot Chris Godsil. – Mojtaba Jazaeri Apr 27 '14 at 22:12
up vote 4 down vote accepted

To give an explicit realization we need to give the group $G$ and a connection set $C$. I will specify the group and explain how to choose the connection set.

Let $\mathbb{F}$ be a field of order $q$ and for $a$, $b$ in $\mathbb{F}$ let $T_{a,b}$ be the map that sends a field element $x$ to $ax+b$. This is invertible if $a\ne0$. Define $G$ to be the set of maps $T_{a,b}$, where $a$ is a non-zero square and $b$ is arbitrary. Then $G$ acts regularly on the edges of $K_q$, and this is our group. (This is where the condition $q\equiv3$ mod 4 comes in.)

The connection set can be taken to be the set of maps $T_{a,b}$ that send the edge $\{0,1\}$ to an overlapping edge, so it consists of the stabilizer of $0$, the stabilizer of $1$, the maps that send $0$ to $1$ and their inverses. (Note that we have two actions for $G$, one on the vertices and one on the edges, and in the least part of the previous sentence I am referring to the vertex action.)

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.