Problem: Let $M$ and $N$ be perfect matchings of a finite graph $G$. Prove that their symmetric difference is a union of disjoint even cycles.
My attempt:
Firstly, if $M = N$, we are done.
Now, let us consider the case when $M \neq N$. Since $M$ and $N$ are perfect matchings of $G$ then $d(v) \ge 2$ for any $v \in V$. Let $e = (u,v)$ be the edge in $M \setminus N$. Since $N$ is a perfect matching then there exists a vertex $v' \in N$ such that $(v,v') \in N\setminus M$. If we keep doing like this, we will obtain a cycle. But, I do not know how to write it down. I hope you could help me with this.