Let G$G$ be a 5$5$-regular graph with Kappa(G) = 2$\kappa(G) = 2$. Prove that Lambda(G) <= 4$\lambda(G) \leq 4$.
As part of my revision for a graph theory I'm doing through some provided questions and answers, however the answer to the above wasn't provided.
I know that a 5$5$-regular graph contains vertices all with degree 5$5$.
So using Whitney's theorem we have: Kappa(G) = 2 <= Lambda(G) <= 4 <= Delta(G) = 5$\kappa(G) = 2 \leq \lambda(G) \leq 4 \leq \Delta(G) = 5$.
I'm just not really sure how to approach proving that Lambda is less than 5$5$.
If every vvertex $v$ has degree 5 then$5$ and Kappa(G) = 2$\kappa(G) = 2$ then there must be 2$2$ internally disjoint paths between any u$u$ and v$v$ in the graph.
because of these two internally disjoint paths, the edges of u$u$ must be split between these two paths, as such one path would have 2$2$ edges and another 3$3$ edges. This would be the same as v$v$, which would at most have 2$2$ edges from one path and 3$3$ from the other.
This means that to destroy the connectivity of the graph, you could cut the two edges from u$u$ and the two edges to v$v$. If one or both of the paths have only 1$1$ edge, then the number of edges to cut will always be <= 4$\leq 4$.
Is this a sufficent proof? Is there anyway of making it simpler/neater?
Thanks
