# Show that the Petersen graph is vertex transitive

Show that the Petersen graph is vertex transitive.

This is my first opinion to show that the Petersen graph is vertex transitive :

The vertices of the petersen graph $G$ are labeled with 2-element subsets of {$1,2,3,4,5$} and two vertices are adjacent if and only if their intersection is empty. This is the Kneser graph labelling (1955). So that $S_5$ is contained in Automorphisms and so Graph $G$ is vertex transitive. In fact automorphism of $G$ = $S_5$.

High appreciated if someone can explain it clearly.

• You can also draw the Petersen graph and provide explicit maps, that map every vertex onto every other. This approach, too, can have an interesting effect: you see vertices move. Your approach above includes (for me at least) that the Petersen graph is isomorph to the construction process described. – Moritz Oct 31 '15 at 8:36
• @Moritz can you explain to me with another way, sorry i still can not show that Petersen graph is vertex transitive. I appreciate any answer, guidance or reference. Thank you – user273952 Nov 1 '15 at 16:36

Consider the Petersen graph in the image on the left. I have labeled the vertices via sets, i.e., $12$ is $\{1,2\} = \{2,1\}$ and $34$ is $\{3,4\} = \{4,3\}$ et cetera. If you apply the permutation $(1,4,5,2,3)$ to the vertices, you get a clockwise rotation, that is, $(1,4,5,2,3)$ takes $\{1,2\}$ and maps it to $\{4,3\}$ and so on. The cycle notation $(1,4,5,2,3)$ means $1 \mapsto 4$ and $4 \mapsto 5$ and $5 \mapsto 2$ and $2\mapsto 3$ and $3\mapsto 1$. All you have to do now is the following: If you find a permutation which maps the inner $5$ vertices onto the outer $5$ vertices, the you are finished because then you can mix the two permutations to map every vertex onto every other. Still unclear?
1. Show that, for any permutation $\pi$ of the set $\{1,2,3,4,5\},$ there is an isomorphism of the Petersen graph which maps each vertex $\{x,y\}$ to $\{\pi(x),\pi(y)\}.$
2. Show that, for any two vertices $\{x,y\}$ and $\{u,v\}$ of the Petersen graph, there is a permutation $\pi$ of $\{1,2,3,4,5\}$ such that $\pi(x)=u$ and $\pi(y)=v.$