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? 



Cayley graph which is isomorphic to the line graph of a complete graph
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.)


