Automorphisms of Lubotzky–Phillips–Sarnak graphs

Question

For the Lubotzky–Phillips–Sarnak (LPS) graph $X^{p,q}$, what is its automorphism group? These graphs are not just ($p+1$)-regular but are Cayley graphs for $G=\mathrm{PSL}_2(\mathbb{F}_q)$, so clearly $G\le \mathrm{Aut}(G)$. But is there anything else known about the structure of the full automorphism group in this case? (As far as I can tell, the necessary condition of this post appears to be true here but the papers cited giving sufficiency do not apply – but I could be mistaken on this point.)

Answer 1

The automorphism group is strictly larger than $G$. Note that the automorphism group is a semidirect product of $G$ with the stabiliser of a single vertex, so it suffices to show that the group of automorphisms fixing a vertex is nontrivial. Indeed, we shall exhibit an order-24 group (isomorphic to $S_4$) of automorphisms fixing a vertex.

The generators of the Cayley graph correspond to the quaternions $a + bi + cj + dk$ of norm $p$ subject to the constraint that $a > 0$ is an odd integer and $b,c,d$ are even integers. (The Jacobi four-square theorem implies that there are $p + 1$ such quaternions.) More specifically, we take the $p + 1$ generators of the Cayley graph to be the images of these quaternions under the homomorphism $\theta$:

$$ \theta(a + bi + cj + dk) := \begin{pmatrix} a + bi & c + di \\ -c + di & a - bi \end{pmatrix} $$

from the group of norm-$p$ Lipschitz integer quaternions to the group $PSL(2,q)$.

Any automorphism of the quaternions which maps generators to generators will induce an automorphism of the Cayley graph which fixes the vertex corresponding to the identity element. There is a group of 24 such automorphisms, namely those of the form:

$$ q \mapsto hqh^{-1} $$

where $h$ is an element of the binary octahedral group. (Although there are 48 elements of the binary octahedral group, $\pm h$ induce the same automorphism, so there are only 24 distinct automorphisms, forming a group isomorphic to the orientation-preserving symmetries of a regular octahedron, which is in turn isomorphic to $S_4$.)

Consequently, we can strengthen your statement to say that $Aut(G) \geq G \rtimes S_4$.