Here is the solution which I am not getting, I am writing it in points along with the doubts I have (written in bold)-:
$1$ Suppose that the Petersen graph has a cycle C of length $7$.Since any two vertices of C are connected by a path of length at most $3$ on $C$ , any additional edge with endpoints on C would create a cycle of length at most $4$ Hence the third neighbor of each vertex on C is not on C.
$2$.Thus there are seven edges from $V \left(C\right) $ to the remaining three vertices.
$3$ By the pigeonhole principle, one of the remaining vertices receives at least three of these edges. This vertex x not on C has three neighbors on C.
$4$.For any three vertices on C, either two are adjacent or two have a common neighbor on C (again the pigeonhole principle applies). Using x, this completes a cycle of length at most 4. We have shown that the assumption of a $7$ cycle leads to a contradiction
In short I am not getting major of it .. I know it is duplicate of here please help me out !