On generators of automorphism group of symmetric graph

Suppose I have a connected graph $G$ that is vertex-transitive. Fix a vertex $v$ in $G$ and let $h$ be an automorphism of graph $G$. If $v$ is adjacent with $h(v)$ and $G$ is symmetric, I need to show that $h$ and the stabilizer of $v$ generate $Aut(G)$.

This problem is from Algebraic Graph Theory by Norman Biggs.

I'm having a brain dead. Attempted to show it directly with connectivity but failed. Used the symmetric property but couldn't cover all cases.

Any help is appreciated. Thanks.

Let $A = {\rm Aut}(G)$, let $A_v$ be the stabilizer of $v$ in $A$, and let $B = \langle A_v, h \rangle$. We have to show that $A=B$. Since $A_v \le B$, it is sufficient to prove that $B$ acts transitively on vertices. In other words, we have to show that all vertices lie in the orbit $v^B$ of $v$ under $B$.
Let $N(v)$ be the set of vertices adjacent to $v$. Then, since $G$ is symmetric, $G_v$ acts transitively on $N(v)$. Since $h(v) \in N(v)$, we have $N(v) \subseteq v^B$.
Now, since $G$ is connected, for any vertex $w$, there is a sequence $v_0=v,v_i,\ldots,v_n=w$ of vertices in which $v_i,v_{i+1}$ are adjacent for all $i$. We show by induction on $n$ that $w \in v^B$. This is clear for $n=0$ and we proved it for $n=1$ in the previous paragraph. By induction we may assume that $v_{n-1} \in v^B$. So there exists $g \in A$ with $v^g=v_{n-1}$. Then $A_{v_{n-1}} = gA_vg^{-1} \le B$.
Now, since $v$ and $h(v)$ are adjacent in $G$, so are $g(v)=v_{n-1}$ and $gh(v)$. Then, since $G$ is symmetric, there exists $f \in A_{v_{n-1}} \le B$ with $fgh(v) = v_n$, so $w = v_n \in v^B$, which completes the induction.
• Each stabilizer is in $B$. What about the automorphisms that fix no vertex? – user297210 Dec 9 '15 at 10:44