# Graph factorization

Theorem by Kano, Lee, and Suzuki:

Every connected cubic bipartite simple graph has a $\{C_n:n\ge 6\}$-factor.

If I have a graph, say $G$, that has $6k$ vertices and satisfies the assumption of the theorem, can I just say that $G$ has $C_6$-factor?

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I think this paper is the same as the one at this link which defines the concept as having an $S$-factor if there is a spanning subgraph whose components come from the set $S$. Thus the theorem states that every graph has there is a spanning subgraph which contains only cycles of length 6 or more, not necessarily exactly 6.
If you look at reference [4] in the paper you can see their figure 1 which is a graph with 48 vertices which can be seen to not be able to be decomposed into 8 copies of $C_6$. With some more practise you will be able to manipulate cubic graphs like this and the one in your question yesterday, but it takes some experimentation...
Does it mean that i cannot use the theorem to say that the graph has $C_6$-factor? –  kim_kibun Feb 20 '13 at 8:58