Petersen graph is tripartite Please help me prove Petersen graph is tripartite.
I know that is not bipartite by brute force, however I'm not sure how toextend this for Tripartite. I have a only clue that The chromatic number of the Petersen graph is 3 and it has 5 cycle.
Do we have any standard proof or do we have to assume things inorder to prove this. kindly advise.
 A: Use "This drawing with order-3 symmetry"

(colors added to figure given here)
A: The Petersen graph can be defined as the graph whose vertices are the two-element subsets of $\{1,2,3,4,5\},$ with two vertices being adjacent if they are disjoint sets.
The Petersen graph is not $2$-colorable (bipartite) because it contains a $5$-cycle, e.g., the cycle
$$\{1,2\},\{3,4\},\{5,1\},\{2,3\},\{4,5\},\{1,2\}.$$
We can show that the Petersen graph is $3$-colorable (tripartite) by exhibiting a proper $3$-coloring of the vertices:
Color a vertex red if its least element is $1$; i.e., the red vertices are
$$\{1,2\},\{1,3\},\{1,4\},\{1,5\}.$$
Color a vertex white if its least element is $2$; i.e., the white vertices are
$$\{2,3\},\{2,4\},\{2,5\}.$$
Color the remaining vertices blue; i.e., the blue vertices are
$$\{3,4\},\{3,5\},\{4,5\}.$$
A: Since the chromatic number is three, i.e., vertices can be colored with three colors such that there is no edge between the same color vertices. 
Now consider a tri-partition of vertex set, where each partition consists of verices of same color. 
