induced subgraph of $G_I^{(u,v)}$ it follows that $w_{j}$ and $w_{{j+1}}$ are also connected in $G_I^{(u,v)}$. Finally, since $w=w_{1}$ and $w'=w_{r}$. It follows that $G_I^{(u,v)}$ is connected for all $u,v\in G(I)$ with $\deg(\lcm(u,v))>d+1$. If $\deg(\lcm(u,v))\leq d+1$, then $G_I^{(u,v)}$ is obviously connected.

\[connected\] Let $I$ be a monomial ideal generated in degree $d$. Then $I$ is linearly related if and only if for all $u, v\in G(I)$ there is a path in $G^{(u,v)}_I$ connecting $u$ and $v$.

Because of Proposition \[graph\] it suffices to show that the following statements are equivalent:

1.  $G^{(u,v)}_I$ is connected for all $u, v\in G(I)$;

2.  for all $u, v\in G(I)$, there is a path in $G^{(u,v)}_I$ connecting $u$ and $v$.

(i)(ii) is obvious.

(ii)(i): Let $w\in G^{(u,v)}_I$ with $w\neq u$. It is enough to show that $w$ is connected to $u$ by a path in $G^{(u,v)}_I$. By assumption $w$ is connected to $u$ by a path in $G^{(u,w)}_I$. Since $w\in G^{(u,v)}_I$ it follows that $\lcm(u,w)$ divides $\lcm(u,v)$. This implies that $G^{(u,w)}_I$ is an induced subgraph of $G^{(u,v)}_I$. Thus the path connecting $w$ with $u$ in $G^{(u,w)}_I$ also connects $w$ and $u$ in $G^{(u,v)}_I$.

Powers of edge ideals of index $>1$
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Let $\MM$ be the set 