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Finite $s$-geodesic transitive graphs under certain girths

Published 6 Jun 2025 in math.CO | (2506.05803v1)

Abstract: For an integer $s\geq1$ and a graph $\Gamma$, a path $(u_0, u_1, \ldots, u_{s})$ of vertices of $\Gamma$ is called an {\em $s$-geodesic} if it is a shortest path from $u_0$ to $u_{s}$. We say that $\Gamma$ is {\em $s$-geodesic transitive} if, for each $i\leq s$, $\Gamma$ has at least one $i$-geodesic, and its automorphism group is transitive on the set of $i$-geodesics. In 2021, Jin and Praeger [J. Combin. Theory Ser. A 178 (2021) 105349] have studied $3$-geodesic transitive graphs of girth $5$ or $6$, and they also proposed to the problem that to classify $s$-geodesic transitive graphs of girth $2s-1$ or $2s-2$ for $s=4, 5, 6, 7, 8$. The case of $s = 4$ was investigated in [J. Algebra Combin. 60 (2024) 949--963]. In this paper, we study such graphs with $s\geq5$. More precisely, it is shown that a connected $(G,s)$-geodesic transitive graph $\Gamma$ with a nontrivial intransitive normal subgroup $N$ of $G$ which has at least $3$ orbits, where $G$ is an automorphism group of $\Gamma$ and $s\geq 5$, either $\Gamma$ is the Foster graph and $\Gamma_N$ is the Tutte's $8$-cage, or $\Gamma$ and $\Gamma_N$ have the same girth and $\Gamma_N$ is $(G/N,s)$-geodesic transitive. Moreover, it is proved that if $G$ acts quasiprimitively on its vertex set, then $G$ is an almost simple group, and if $G$ acts biquasiprimitively, the stabilizer of biparts of $\Gamma$ in $G$ is an almost simple quasiprimitive group on each of biparts. In addition, $G$ cannot be primitive or biprimitive.

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