Directed Graphs on Relations - Set Theory These questions were from an assignment I had some time ago but the solutions were not provided.
A permutation $P$ on a finite set $A$ is a binary relation with the property that for each $a∈A$, there is precisely one element of $P$ of the form $(a, x)$, and precisely one element of $P$ of the form $(x, a)$. 
For example, $\{(1, 4), (2, 2), (3, 1), (4, 3)\}$ is a permutation of the set $\{1, 2, 3, 4\}$.
Q: If $P$ is a permutation, describe the general form of its directed graph diagram.
My solution: The general form of its directed graph being that it is one-to-one is that each element will have exactly one arrow going from it and one to it. 
(The marker of this assignment states, "What does this mean for the structure of the diagram?").
Q: If $P$ and $Q$ are permutations on the set $\{1, 2, ... n\}$, how is the directed graph diagram of $P$ related to the directed graph $Q^{-1} \circ P \circ Q$? A: I claimed that they will have the exact same figure, meaning that the shape of the closed circuits will always be the same. Apparently this is correct according to the marker, but I am not sure how to prove this statement.
A: For your first question note that you will have disjoint circuits in your directed graph. 
For your second question, you will get the same figure (though the corresponding permutation may be different from $P$ or $Q$). This follows from a result proved here.
A: *

*The marker might have been looking for you to say that the directed graph diagram of the permutation will consist of some number of disjoint cycles.


*As you stated, the directed graph diagram of $Q^{-1} \circ P \circ Q$ will have the exact same shape as the directed graph diagram of $P$, but nodes in the cycles might change.
This basically follows from the fact that $( Q^{-1} \circ P \circ Q ) ( Q^{-1} ( a ) ) = Q^{-1} ( P ( a ) )$, and so the cycle $$a_1 \rightarrow a_2 \rightarrow \cdots \rightarrow a_n \rightarrow a_1$$ in $P$ will correspond to the cycle $$Q^{-1} (a_1) \rightarrow Q^{-1} (a_2) \rightarrow \cdots \rightarrow Q^{-1} ( a_n ) \rightarrow Q^{-1} (a_1)$$ in $Q^{-1} \circ P \circ Q$.
