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Let $G=(V,E)$ be a finite simple $k-$regular graph ($k\geq 1$). Does $G$ necessarily contain a subset $E'\subset E$ of edges such that only isolated edges and cycles occur as connected components in $(V,E')$?

(The answer is easily yes for $k=1,2$.)

A counterexample would easily give a counterexample to question "Antipodal" maps on regular graphs? in the case $D=2$ by considering the complementary graph of $G$ (respectively of two disjoint copies of $G$ if $G$ is "too small").

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Roland Bacher
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1 Answers1

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It seems like such a subset should always exist.

Consider the bipartite graph on $2|V(G)|$ vertices corresponding to the adjacency matrix of $G$. Since this graph is regular, by Hall's Theorem it has a perfect matching. In terms of the original $G$, this corresponds to a permutation $\sigma$ on $V(G)$ such that $v$ and $\sigma(v)$ are always adjacent. The cycles of $\sigma$ would then give you the desired decomposition.

Kevin P. Costello
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