Let $G$ a graph and $M_1 , M_2$ not equal perfect matchings, show that every connected component of $G[M_1 \triangle M_2]$ is a cycle. Where $G[M_1 \triangle M_2]$ is the induced graph by the symmetric difference of these two perfect matchings.
What I've got so far is that since $M_1, M_2$ are not equal then $M_1 \triangle M_2 \neq \emptyset$ and $G$ has an even number of vertices, also that the induced graph has the same number of vertices than $G$.
Should I suppose a contradiction taking a connected component? Any hints?
Take the multigraph union $G[M_1\cup M_2]$ of the two matchings $M_1$ and $M_2$.
Since a perfect matching saturates all vertices, the multigraph union results in each vertex having degree 2.
It is a well-known result that if every vertex in a connected component is of degree 2, the component is a cycle. Therefore, every component in this graph is a cycle.
Next, we note that $G[M_1 \triangle M_2]$ can be obtained by removing edges of multiplicity 2 from $G[M_1\cup M_2]$.
The result follows.