Statement: If a graph has more than one perfect matchings, then it has an even cycle.
I was trying to prove this result. My try:
Let $M$ and $M'$ be two perfect matchings in graph $G$. Symmetric difference of $M$ and $M'$ consists of vertices of degrees 0 or 2. Therefore components of the symmetric difference are either isolated vertices or cycles.
But I am stuck here. How do I prove that at least one of those cycles must be an even cycle? Any help. Thank You.
Firstly, don't forget to show that there must be some degree 2 vertices in that symmetric difference, and hence at least one cycle.
Lastly, show that the edges of any such cycle must alternate between $M$ and $M'$, making it even length.