In my textbook, the author said “we see that the automorphism group of the Petersen graph has order at least 120, and therefore it is at least 3-arc transitive.”
I know the automorphism group of the Petersen graph is exact $S_5$, but I don’t know why the words of author can follow.
Can someone help me about this? Thanks a lot!
Let $G$ be the automorphism group and $X$ the set of $3$-arcs of the Petersen graph. We know that $|G|\ge120$. Because the Petersen graph has girth $5$, we can easily count $|X|=10\cdot3\cdot2\cdot2=120$. Suppose $g\in G$ fixes $x=(x_1,x_2,x_3,x_4)\in X$. Then $g$ must be the identity: Because the Petersen graph has diameter $2$, $x_1$ and $x_4$ share a neighbor $x_5$ that $g$ must fix, and then each remaining vertex is adjacent to a unique vertex in this fixed cycle and must therefore also be fixed. Then $|G\cdot x|=|G|/|G_x|\ge120=|X|$, so $G$ acts transitively on $X$.