For any graph $G$ let $\eta(G)$ be the [Hadwiger number](http://en.wikipedia.org/wiki/Hadwiger_number) of $G$.


Is there for every graph $G$ a graph $2G$ such that


-- $\chi(2G) = 2\chi(G)$, and


-- $\eta(2G) = 2\eta(G)$?


For each one of the above conditions it is easy to construct a graph $2G$ to $G$ such that the condition holds, but I haven't been able to double both the coloring number and the Hadwiger number at the same time with a "universal construction". 



Graph construction to double coloring & Hadwiger number
Define $K\_n'$ to be the graph obtained from the complete graph on $n$ vertices by subdividing each edge once. Let $G$ be a graph with $\chi(G)=c$ and $\eta(G)=h$. Define $2G$ to be the disjoint union of $K\_{2h}'$ and $K\_{2c}$. Assuming [Hadwiger's conjecture](http://en.wikipedia.org/wiki/Hadwiger_conjecture_%28graph_theory%29), we have $c \leq h$, and so $\eta(2G)= 2 \eta(G)$ and $\chi(2G)=2\chi(G)$ (since $K\_n'$ is 2-colourable for all $n$). 


Of course, this may not be the type of construction you had in mind, but it works (assuming Hadwiger's conjecture is true). 


