[Paper Review] A note on the automorphism groups of Johnson graphs

S. Morteza Mirafzal|arXiv (Cornell University)|Feb 8, 2017

Finite Group Theory Research9 references5 citations

TL;DR

This paper provides an elementary proof that the automorphism group of the Johnson graph $J(n,i)$ is isomorphic to $\mathrm{Sym}(n)$ when $n \neq 2i$, and to $\mathrm{Sym}(n) \times \mathbb{Z}_2$ when $n = 2i$, confirming a conjecture from previous work. The proof uses distance-based vertex structure analysis and group action arguments without heavy group-theoretic machinery.

ABSTRACT

The Johnson graph $J(n, i)$ is defined as the graph whose vertex set is the set of all $i$-element subsets of $\{1, . . ., n \}$, and two vertices are adjacent whenever the cardinality of their intersection is equal to $i$-1. In Ramras and Donovan [SIAM J. Discrete Math, 25(1): 267-270, 2011], it is proved that if $ n eq 2i$, then the automorphism group of $J(n, i)$ is isomorphic with the group $Sym(n)$ and it is conjectured that if $n = 2i$, then the automorphism group of $J(n, i)$ is isomorphic with the group $ Sym(n) imes \mathbb{Z}_2$. In this paper, we will find these results by different methods. We will prove the conjecture in the affirmative.

Motivation & Objective

• To resolve a conjecture regarding the automorphism group of Johnson graphs $J(n,i)$ when $n = 2i$.
• To provide an elementary proof—using only basic group theory and graph structure—of the automorphism group for Johnson graphs.
• To establish that the automorphism group is $\mathrm{Sym}(n)$ when $n \neq 2i$, and $\mathrm{Sym}(n) \times \mathbb{Z}_2$ when $n = 2i$, via distance-transitive properties and stabilizer analysis.
• To show that the complementation map $v \mapsto v^c$ is an automorphism not in the symmetric group action, and that it commutes with all symmetric group automorphisms.

Proposed method

• Analyzes the distance partition $\Gamma_i(x)$ from a fixed vertex $x$ to characterize vertices via intersections of neighborhoods.
• Uses the fact that $\Gamma_1(x)$, the neighborhood of $x$, is isomorphic to the line graph $L(K_{m,n-m})$ to analyze the local automorphism structure.
• Applies the isomorphism $\mathrm{Aut}(L(K_{m,n-m})) \cong \mathrm{Aut}(K_{m,n-m})$ to bound the size of the stabilizer subgroup $G_v$.
• Shows that the complementation map $\alpha(v) = v^c$ is an automorphism of order 2 not contained in the image of $\mathrm{Sym}(n)$, and commutes with all $f_\theta$.
• Proves that the group generated by $\mathrm{Sym}(n)$ and $\alpha$ forms a direct product $\mathrm{Sym}(n) \times \mathbb{Z}_2$ when $n = 2m$, and that this group achieves the upper bound on $|\mathrm{Aut}(J(n,m))|$.
• Uses the orbit-stabilizer theorem and vertex transitivity to relate $|\mathrm{Aut}(J(n,m))|$ to $|G_v| \cdot \binom{n}{m}$, leading to the final group isomorphism.

Experimental results

Research questions

• RQ1 What is the automorphism group of the Johnson graph $J(n,i)$ when $n = 2i$?
• RQ2 Is the automorphism group of $J(n,i)$ isomorphic to $\mathrm{Sym}(n) \times \mathbb{Z}_2$ when $n = 2i$, as conjectured?
• RQ3 Can the automorphism group of $J(n,i)$ be determined using only elementary group-theoretic and graph-theoretic methods?
• RQ4 Why does the complementation map $v \mapsto v^c$ act as an automorphism not in the image of $\mathrm{Sym}(n)$, and how does it interact with the symmetric group action?

Key findings

• When $n \neq 2m$, the automorphism group of $J(n,m)$ is isomorphic to $\mathrm{Sym}(n)$, as the group order is bounded by $n!$ and the symmetric group action achieves this bound.
• When $n = 2m$, the automorphism group of $J(n,m)$ is isomorphic to $\mathrm{Sym}(n) \times \mathbb{Z}_2$, with the $\mathbb{Z}_2$ factor generated by the complementation map $\alpha(v) = v^c$.
• The complementation map $\alpha$ is an automorphism of order 2 and is not in the image of the symmetric group action $\mathrm{Sym}(n)$, as shown by contradiction using fixed points and transpositions.
• The automorphism group $\mathrm{Aut}(J(n,m))$ is the direct product $H \times \langle \alpha \rangle$, where $H \cong \mathrm{Sym}(n)$ is the image of the symmetric group action and $\alpha$ commutes with all $f_\theta$, so $H \triangleleft G$ and $\langle \alpha \rangle \triangleleft G$.
• The stabilizer subgroup $G_v$ has size at most $|\mathrm{Aut}(K_{m,n-m})|$, and this bound is tight, leading to the conclusion that $|\mathrm{Aut}(J(n,m))| \leq 2(2m)!$ when $n = 2m$, matching the order of $\mathrm{Sym}(2m) \times \mathbb{Z}_2$.
• The group $G = \mathrm{Aut}(J(n,m))$ is equal to $H \langle \alpha \rangle$, and since $f_\theta \alpha = \alpha f_\theta$ for all $\theta$, the product is direct, confirming $G \cong \mathrm{Sym}(n) \times \mathbb{Z}_2$ when $n = 2m$.