On the automorphism group of a Johnson graph

TL;DR

This paper proves a conjecture about the automorphism group of Johnson graphs, showing it is a specific product of symmetric groups and a complementation map, using elementary group theory and clique analysis.

Contribution

It confirms the conjecture that for n=2i, the automorphism group of Johnson graphs is S_n times the group generated by the complementation map.

Findings

Automorphism group of Johnson graph J(n,i) for n=2i is S_n × <T>.

The proof uses elementary group theory and clique structure analysis.

The conjecture by Ramras and Donovan is resolved affirmatively.

Abstract

The Johnson graph $J(n,i)$ is defined to the graph whose vertex set is the set of all $i$-element subsets of $\{1,\ldots,n\}$, and two vertices are joined whenever the cardinality of their intersection is equal to $i-1$. In Ramras and Donovan [\emph{SIAM J. Discrete Math}, 25(1): 267-270, 2011], it is conjectured that if $n=2i$, then the automorphism group of the Johnson graph $J(n,i)$ is $S_n \times \langle T \rangle$, where $T$ is the complementation map $A \mapsto \{1,\ldots,n\} \setminus A$. We resolve this conjecture in the affirmative. The proof uses only elementary group theory and is based on an analysis of the clique structure of the graph.