# Cayley graph which is isomorphic to the line graph of a complete graph

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?

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Thank you every one who answer this question – Mojtaba Jazaeri Apr 27 '14 at 11:42
See Chris Godsil's answer to mathoverflow.net/questions/150744/… – Tony Huynh Apr 27 '14 at 13:55
Thanks a lot Chris Godsil. – Mojtaba Jazaeri Apr 27 '14 at 22:12

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.)