# Arc transitivity of the complete graph

Recall that a graph $G$ is arc transitive if the natural action of $\mathrm{Aut}(G)$ on $A(G) = \{ (u,v) | \{u,v\} \in E(G)\}$ is transitive.

In other words, given $(u,v),(u'.v') \in A(G)$ one finds a $g \in \mathrm{Aut}(G)$ such that $g (u,v) := (g(u),g(v)) = (u',v').$

Our lecturer said it is not completely straightforward to show that $K_n$ is an arc transitive graph. As far as I can see since $\mathrm{Aut}(K_n) = S_n$ one can always find a suitable permutation that maps a pair of vertices into another pair of vertices thus showing that $K_n$ is indeed arc transitive.

Am I missing something? I recall the lecturer talked about considering some cases (i.e if the arcs are not disjoint) but I don't see how this could create any real obstacle in finding the suitable permutation.

Am I missing something crucial here?

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It might help if you told us what $K_n$ is. Is it the complete graph on $n$ vertices? If so, the problem amounts to showing that $S_n$ acts 2-transitively, which I would agree is not very taxing (assuming $n \ge 2$). – Derek Holt Oct 27 '12 at 13:03

$S_n$ acting transitively on $K_n$ only gives you that for each two vertices $v,v'\in V(K_n)$, there is a $\sigma\in S_n$ so that $v\sigma=v'$. You want to show that for any four vertices $v,w,v',w'\in V(K_n)$ (with $v,w$ distinct, as well as $v',w'$), you can find a $\sigma$ so that $v\sigma = v'$ and $w \sigma = w'$. This means $S_n$ has to be $2$-transitive, which is stronger than just transitive. For example, $D_n$ is transitive but not $2$-transitive for any $n> 3$, which is easy to see visually by its action on the $n$-cycle.