$v/\tilde{w}, u/\tilde{w}\in G({I^{k-1}})$, a contradiction.

Furthermore, $w/\tilde{v}, v/\tilde{v}\in G({I^{k-1}})$, and so if $\deg(\lcm(w,v))>2k+1$, by using the induction hypothesis there exists a path between $w/\tilde{v}$ and $v/\tilde{v}$ in $G^{(w/\tilde{v},v/\tilde{v})}_{I^{k-1}}$. As above this implies that $v$ and $w$ are connected in $G^{(w,v)}_{I^{k}}$ and hence in $G^{(u,v)}_{I^{k}}$, since $G^{(w,v)}_{I^{k}}$ is a subgraph of $G^{(u,v)}_{I^{k}}$. In the case that $\deg(\lcm(w,v))\leq 2k+1$, it is obvious that $v$ and $w$ are connected in $G^{(u,v)}_{I^{k}}$. Also by construction of $w$, we have $\deg(\lcm(u,w))>2k+1$, and the monomials $u$ and $w$ have a common factor, say $\tilde{w}\in G(I)$ such that $w/\tilde{w}, u/\tilde{w}\in G({I^{k-1}})$. Again by using our induction hypothesis, we conclude that there exists a path between $w$ and $u$ in $G^{(u,v)}_{I^{k}}$. Therefore $u$ and $v$ are connected in $G^{(u,v)}_{I^{k}}$ also in this case.

(b)(c) is obvious.

(c)(a): Assume that $G$ is not gap free. Thus there exist four different vertices $i, i', j, j'$ such that $\{i, i'\}, \{j, j'\}\in E(G)$ while the edges $\{i, j \}, \{i, j'\}, \{i', j\}$ and $\{i', j'\}$ are not in $E(G)$. Let $k>0$ be an arbitrary integer. We will show that $u=(x_{i}x_{i'})^k$ and $v=(x_{j}x_{j'})^k$ are not connected in $G_{I^k}^{(u, v)}$. Suppose that they are connected. Then there exists a monomial $w\in 