Find a regular graph in which $\kappa(G) \neq \lambda(G)$ The vertex-connectivity, $\kappa(G)$, is the minimum number of nodes whose deletion from a graph G disconnects it.  

The edge-connectivity of G, $\lambda(G)$, is the minimum cardinal of the cuts in G.  

Now can you find a regular graph in which $\kappa(G) \neq \lambda(G)$ ?

Note: I know this graph is not 3-regular.
The following graph is 4-regular with $\kappa =1$ , and $\lambda =2$. Note that for any graph $G$, $\kappa(G)\leq \lambda(G)$.
The following graph is 4-regular with $\kappa =1$ , and $\lambda =2$. Note that for any graph $G$, $\kappa(G)\leq \lambda(G)$.