If G is a connected 2k-regular graph ,prove that if G has an even number of edges then there is a k-regular subgraph H of G such that V(H)=V(G).
Well here are my thoughts:
Because G is connected and the degree of all vertices is even, G has an Euler circuit. I assume that if we delete every other edge in the circuit then the degree of the vertices will be k. How do we use the fact that G has even numbers of edges and can we say something if g has an odd numbe of edges? There is something that I can't see.