I have been asked to prove the following as an exercise:
"Let $Sun_{2}$ be the graph obtained from $Sun_{4}$ by deleting two vertices of degree $1$ of distance $3$ from each other. Show that every prime $Sun_{2}$-free bipartite graph is the bipartite complement of a regular graph of degree $1$."
Here is $Sun_{4}$:
I'm sure this is false. Take the graph $P_{4}$ as a counter-example. $P_{4}$ is clearly prime, $Sun_{2}$-free, and bipartite, but it is the bipartite complement of only $K_{2}+2K_{1}$ (with the isolated vertices placed in different parts), which clearly is not $1$-regular. $P_{6}$ serves as another counter-example.
I am also appreciating, as I write this, that the bipartite complement operation is not well-defined. If a graph is disconnected and so admits more than one bipartition, each bipartition may result in a different bipartite complement. Could someone please verify this?
However this does not seem to resolve the exercise.
It seems strange to be asked to prove something which is so clearly false. Am I missing something?
Thanks muchly.