Let $G$ be a permutation graph
By definition if $\pi$ is a permutation of $\{1...n \}$ the vertices $i,j$ are connected by an edge if and only if $ i\lt j $ and $\pi_i \gt \pi_j$
The edges of a graph are transitive (ordered) if for every $(i,j) ,(j,k) \in E \implies (i,k) \in E$
Prove the the complement $\hat G$ of $G$ is a permutation graph and $G$ is transitive
What I got: Since the complement $\hat G$ of $G$ contains all the edges which $G$ does not contain.
$i<j$ and $\pi_i \le\pi_j$ then $(i,j) \in \hat G$
which permutation does that correspond to ?
For the second part:
$$i \lt j \quad \pi_i \gt \pi_j$$ $$j \lt k \quad \pi_j \gt \pi_k$$ $$\implies$$
$$i\lt j \lt k \quad \pi_i\gt\pi_j \gt\pi_k$$
so the permutation graph is transitive (because $\pi$ is)
Would appreciate any help with the first part
For 1), take the reverse of the permutation. Note that the condition $\pi_i \le\pi_j$ is equivalent to $\pi_i < \pi_j,$ because $\pi$ is a permutation, and $\pi_i \ne\pi_j$ when $i\ne j.