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I am reading "Algebraic Graph Theory" by Biggs 1974. In the section about symmetric graphs, it is stated that:

A vertex transitive graph $X$ is symmetric, if and only if each vertex-stabilizer $G_v$ is transitive on the set of vertices adjacent to $v$.

I see why this is a necessary condition, but I do not see why this is a sufficient condition. Can anybody give a short proof of this Lemma?

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The book Algebraic Graph Theory by Biggs is very good in this field I'm fan of this book –  Babak Miraftab May 25 '12 at 10:00

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It is an important question because it is related to distance regular graphs. First you assume that each $G_v$ is transitive on $N(v)$. Let $u,v,x$ and $y$ be vertices in graph; such that $u\sim v$ and $x \sim y$. The graph $X$ is vertex transitive, so there exists an automorphism $\pi$ such that $\pi(v)=y$. $1=d(u,v)=d(\pi(u),\pi(v))=d(\pi(u),y)$. By assumption there exist $g \in G_y$ such that $g(\pi(u))=x$ so the graph $X$ is symmetric

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