Let $G$ be a connected graph, with minimal degree at least $2$ and such that it has a complete matching ($1$-factor). Does $G$ necessarily have a $2$-factor (i.e. $2$-regular spanning subgraph)?
I know that $2$-factors are unions of edge-disjoint cycles and such things occur e.g. if the degree of every vertex is even. But in the generality above I cannot even decide whether the statement is true or false.
Not necessarily. Here is a counterexample, with the perfect matching highlighted. In any 2-factor, all of the 3-cycles would have to be included, but then they end up meeting at the vertices of degree 7.
Here's how you come up with such a graph.
We're allowed to have degree-2 vertices, and such a vertex forces both of the incident edges to be used in any 2-factor. If you have a triangle in which two vertices have degree 2, that triangle must be a component of the 2-factor. If you have two such triangles meeting at a vertex, then there can't be a 2-factor.
Now just fit that into a bigger graph which has the other properties you want.
(I didn't really need 3 triangles meeting at a point, I just thought that it looked cooler that way.)