I want to introduce you one of my midterm question in Graph Theory. I could not do this question in my exam and i wonder how to do that.
Prove that every edge in the Petersen Graph belongs to exactly 4 5-cycles, and use this to show that the Petersen Graph has exactly 12 5-cycles.
Girth of a Petersen Graph 5.
(Another question in my mind is below.
We know Petersen graph has 120 automorphism.
Is there any relationship between the being 12 5-cycle and having 120 automorphisms such that a cycle has 2n automorphisms
n : number of vertices in the graph.
So we, have 12 cyle * 5 length of the cycle * 2 = 120 automorphisms ? )
I want to learn the proof of my midterm question. Could you please help ?
Best Answer
It's convenient to work with the Kneser graph interpretation of the Petersen graph: its vertices are the pairs $$\{1,2\}, \{1,3\}, \{1,4\}, \{1,5\}, \{2,3\}, \{2,4\}, \{2,5\}, \{3,4\}, \{3,5\}, \{4,5\}$$ and its edges join the pairs that don't intersect. (For example, $\{1,2\}$ is adjacent to $\{3,4\}$, $\{3,5\}$, and $\{4,5\}$.)
To show that every edge is contained in four $5$-cycles, we prove this for the edge between $\{1,2\}$ and $\{3,4\}$. (This is fine to assume because the Petersen graph is edge-transitive: there is an automorphism sending every edge to this edge.) Then the remainder of the cycle can be chosen as follows:
Altogether, we have two choices for $x$ and two choices for $y$, so the cycle can be chosen in four ways. We can list out the options: they are
Finally, if each of $15$ edges is contained in four $5$-cycles, that gives $15 \cdot 4 = 60$ cycles total; but each cycle is counted $5$ times, once for each of its edges, so the Petersen graph contains $\frac{60}{5} = 12$ cycles.
You also ask if there is a connection between the Petersen graph having $12$ $5$-cycles, a $5$-cycle having $10$ automorphisms, and the Petersen graph having $12 \cdot 10$ automorphisms.
This is certainly not a general rule for all graphs that we can count automorphisms this way. It works for the Petersen graph because it is highly symmetric. We can obtain all $120$ automorphisms as follows:
It is this "can be completed to a unique automorphism" bit that won't work in general. In other graphs, this can fail in a couple of ways: