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Let $G$ be a connected vertex transitive graph and $G_v$ denote stabilizer of the vertex $v$. If $h$ is any automorphism of $G$ for which $d(v,h(v))=1$, and $G$ is symmetric,the $h$ and $G_v$ generate $\operatorname{Aut}(G)$.

We say that $G$ is symmetric if ,for all vertices $u,v,x,y$ of $G$ such that $u$ and $v$ adjacent, and $x$ and $y$ are adjacent there is an automorphism $g$ in $\operatorname{Aut}(G)$ for which $g(u)=x$ and $g(v)=y$

I think we should use this fact: Let $G$ be group acting transitively on set $X$, $H$ be a subgroup of $G$ and $G_a$ be stabilizer of G then $G=HG_a$ if and only if H is transitive.

Please advise me.

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up vote 4 down vote accepted

I wasn't sure what a symmetric graph was, but according to Wikipedia it means the same as arc-transitive. So $G_v$ is transitive on the set of vertices adjacent to $v$. Let $X = \langle G_v, h \rangle$ and let $H$ be the orbit of $X$. Then, since $H$ contains $h(v)$, and $v$ and $h(v)$ are adjacent, so $H$ contains all vertices adjacent to $v$. Also, since the conjugate of $G_v$ by $x \in X$ is $G_{x(v)}$, $X$ contains $G_w$ for all $w \in H$ so now, by a straightforward induction on $n$, $H$ contains all vertices at distance $n$ from $v$, and hence $H=G$ since $G$ is connected. Then, by the result you mention, $X={\rm Aut}(G)$.

Exercise: find an example in which the result is false if we do not assume that $G$ is symmetric.

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