Let $M$ be a matching in a graph $G$ and let $C$ be a cycle in $G$ of length $2k+1$ for some integer $k \geq 1$. Suppose $C$ contains exactly $k$ edges of $M$, and has one vertex $x$ that is not incident with an edge of $M$. Prove that $M$ is maximum in $G$ if and only if $M'$ is maximum in $G'$, where $M' = M \backslash E(C)$ and $G'$ is the graph obtained from $G$ by contracting the edges of $C$.
I wanted to confirm if this statement correct since I seem to have come up with a counterexample. We can take $C$ to be a triangle (3-cycle) and add an extra edge $e$ to the triangle to get $G$. We can then take $M$ to be any one of the two edges connected to $e$ so that the ocnditions of the theorem are satisfied. But this gives $G' = e$ and $M'$ is the empty graph which is clearly not maximal in $G'$.