Prove that if P and Q are permutation matrices with $(P-I)(Q−I) = 0$ then they represent disjoint permutations. Before even starting let me make clear this question is not duplicate of this which asks for proving just the inverse statement.

Prove that if $P$ and $Q$ are permutation matrices with $(P-I)(Q−I)=0$ then, they represent disjoint permutations

MY TRY :- Let P and Q be the matrices corresponding to the respective permutations $p$ and $q$ in cycle notation. Let $p$ and $q$ do not represent disjoint permutations. For e.g. $p = (123)$ and $q=(345)$ We have that $$
P =
\begin{pmatrix}
0 & 1 & 0 &  0 & 0 \\
0 & 0 & 1 &  0 & 0\\ 
1 & 0 & 0 &  0 & 0\\
0 & 0 & 0 &  1 & 0\\
0 & 0 & 0 &  0 & 1
\end{pmatrix} \text{ and } Q = \begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0\\ 
0 & 0 & 0 & 0 & 1\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}.$$ Now $PQ$ being a permutation will not have two $1$'s in same row. But $P+Q-I$ have two $1$'s and one $-1$ in the "third" row, and $PQ\neq P+Q-I$ which is equivalent $(P-I)(Q-I)\neq 0$ so we reach a contradiction visually. But isn't there some clear method to prove it theoritically?
I am new to group theory. Please ask for clarifications in case of any discrepancies.
 A: Hint: Let $\pi$ denote the permutation corresponding to $P$, and let $e_1,e_2,\dots,e_n$ denote the standard basis of $\Bbb F^n$. Suppose that $\pi = \sigma_1\cdots \sigma_k$ is a decomposition into disjoint cycles (including all "cycles with length $1$"). Let $S_k = \{a_1,\dots,a_\ell\}$, where $\sigma_k = (a_1 \cdots a_\ell)$.
Note that $\ker(P - I)$ is spanned by the vectors $v_k = \sum_{j \in S_k} e_j$. In particular, if $S_k = e_p$, then $e_p \in \ker (P-I)$.  Because $P$ is orthogonal, we have $\operatorname{im}(P - I) = \ker(P - I)^\perp$.
A: The assumption that $PQ = P + Q - I$ is [a priori] stronger than we need.  We may instead simply work with the assumption that the diagonal of $P+Q-I$ only has entries in $\{0,1\}$.
If we let $1 \leq k \leq n$ be arbitrary and write $A_{kk}$ for the $k$th diagonal entry of $A$, then $$\begin{align*}(P+Q)_{kk}-1 = (P&+Q-I)_{kk} \in \{0,1\} \\ &\iff (P+Q)_{kk} \in \{1,2\} \\ &\iff P_{kk} = 1 \text{ or } Q_{kk} = 1\end{align*}$$
where the justification for the last $\implies$ is due to $P_{kk}, Q_{kk} \in \{0,1\}.$
Now we're basically finished.  That the permutation matrices $P$ and $Q$ represent disjoint permutations is equivalent to the statement that, for any $1\leq k\leq n,$ $$P_{kk} = 0 \implies Q_{kk} = 1 \\\text{ and } \\Q_{kk} = 0 \implies P_{kk} = 1,$$ which is logically equivalent, since $P_{kk}, Q_{kk} \in \{0,1\}$, to $$P_{kk} = 1 \text{ or } Q_{kk} = 1$$

A final note: If $P,Q$ are permutation matrices and the diagonal of $P+Q-I$ has entries in $\{0,1\}$, then we can show $PQ = QP = P+Q-I,$ so this is not actually a strengthening/weakening of the theorem.
A: "$P$ and $Q$ are disjoint" means that for every $i$,
$$
Pe_i \ne e_i \implies Qe_i = e_i\\
Qe_i \ne e_i \implies Pe_i = e_i
$$
but
$$
Qe_i = e_j \ne e_i \implies 0 = (P-I)(Q-I)e_i=(P-I)(e_j-e_i) \\
\implies P(e_j-e_i) = e_j-e_i\implies Pe_i = e_i
$$
and
$$
Pe_i = e_j \ne e_i \implies 0 = (P-I)(Q-I)Q^Te_i=(P-I)(e_i -Q^Te_i) \\
\implies P(e_i-Q^Te_i) = e_j - PQ^Te_i = e_i-Q^Te_i\implies e_i = Q^Te_i\\
\implies Qe_i = e_i
$$
