Prove that every 8-regular graph has 4-and 2-regular spanning subgraph!

Question

Prove that every $8$-regular graph has $4$- and $2$-regular spanning subgraphs.

Note: A graph is spanning subgraph, if it contains every vertex of the original graph. Furthermore this example's from Hamiltonian cycle/path topic.

Answer 1

This is part of a general result called Petersen's 2-factor theorem, which is not too difficult to prove as a whole.

Theorem (Petersen): Let $G$ be a $2k$-regular graph. Then $G$ has a decomposition into $k$ edge-disjoint $2$-factors (where a $2$-factor is a $2$-regular spanning graph).

Proof: Without loss of generality, let $G$ be connected. Since $G$ has only vertices of even degree, it follows that $G$ has an Euler cycle. Take an Euler cycle of $G$ and define a corresponding orientation $D(G)$ according to the cycle.

Now, split each vertex $v$ of $G$ into two distinct vertices colored red and blue, say $v_r$ and $v_b$. Let us now define another (undirected) graph $G'$ with vertex set $V_r \times V_b$. For each directed arc $(u,v)$ in our digraph $D(G)$, draw the edge $(u_r,v_b) \in E(G')$. This makes $G'$ into a $k$-regular bipartite graph.

By Hall's marriage theorem, $G'$ has a decomposition into $k$ edge-disjoint perfect matchings. If we now identify $v_r$ and $v_b$ for each vertex, the previously defined perfect matchings gives us a decomposition of $G$ into $k$ edge-disjoint $2$-factors, as required. $\square$

Your result now follows easily from Petersen's theorem. Take any $2$-factor to get your $2$-regular spanning graph. Take any two $2$-factors to get a $4$-regular spanning graph.

In general, for any $2k$-regular graph $G$, there exists $2r$-regular spanning graphs of $G$ for any $0< r \le k$.