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By the Euler characteristic and double counting edges with the cubic assumption, we get $v=2f-4$. So v is even, and it suffices to prove that f is even too. By the defining property of $\mathcal{G}$ and double counting edges, $\sum_{k \equiv 0 mod 4} k f_k=2e$, where $f_k$ is the number of faces of degree $k$. Reducing mod 4, we get that $2e \equiv 0 mod 4$, ...

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