I'm seeking a simple graph $G$ of the following type:


* It contains two disjoint copies of $K\_6$ (the complete graph on 6 nodes), $H$ and $H'$ say.
* Any one-factor of $G$ must contain either (a) a one factor of $H$ and no edges in $H'$ or (b) a one factor of $H'$ and no edges in $H$.
* There exists a one-factor of $G$ that contains a one-factor of $H$.
* There exists a one-factor of $G$ that contains a one-factor of $H'$.



> 
> **Question**: Does $G$ exist?
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I'm also interested in the same problem with $K\_6$ replaced by $K\_{2n}$.


My motivation for this question comes from an attempt to rephrase a question about Latin squares as a question about one-factors.



A graph $G$ with two $K_6$ subgraphs, in which any one-factor of $G$ induces a one-factor in exactly one of the $K_6$ subgraphs?
Here is a proof that $G$ does not exist. Let $M\_1$ be a (red) 1-factor which contains a 1-factor $M\_1'$ of the first $K\_6$, and $M\_2$ be a (blue) 1-factor which contains a 1-factor $M\_2'$ of the second $K\_6$. Note that $M\_1 \triangle M\_2$ is a union of even cycles, with alternating red and blue edges. Observe that we can get a new 1-factor by 'flipping' the edges on any such cycle. Therefore, there must be a cycle $C$ of $M\_1 \triangle M\_2$ which contains all the edges of $M\_1' \cup M\_2'$, otherwise we violate the second condition. Let $e \in E(C) \cap E(M\_1')$. Traverse $C$ clockwise and let $f$ be the next edge of $C$ which is also in $E(M\_1')$. Let $P$ be the subpath of $C$ beginning at $e$ and ending at $f$. Note that there is an edge $g$ between the ends of $P$ because of the first $K\_6$. Therefore, if we flip the edges of $M\_1$ along the even cycle $P \cup g$, we get another 1-factor. However, the intersection of this 1-factor with the first $K\_6$ is neither empty nor a 1-factor. 


Note that this proof works for any $K\_{2n}$ for $n \geq 2$. 


