Prove or disprove that $PQ = P + Q - I$ if $P$ and $Q$ are disjoint permutation matrices whose cycle lengths sum to $n.$ 
Prove or disprove that if the matrices $P$ and $Q$ represent disjoint permutation cycles in $S_{n}$ with sum of cycle lengths equal to $n,$ then $PQ = P+Q-I$.

MY TRY:
Let's start by an example. Let $P$ and $Q$ be the matrices corresponding to the respective permutations $p = (1 \, 2)$ and $q = (3 \, 4 \, 5)$ in cycle notation. We have that $$
P =
\begin{pmatrix}
0 & 1 & 0 &  0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 &  0 & 0\\ 
0 & 0 & 0 &  1 & 0\\
0 & 0 & 0 &  0 & 1
\end{pmatrix} \text{ and } Q = \begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0\\ 
0 & 0 & 0 & 0 & 1\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}.
$$ It seems obvious that the matrix $PQ$ representing the permutation $pq = (1 \, 2)(3 \, 4 \, 5)$ will be $P+Q-I,$ as the "untouched" $1$s in the matrices are simply canceled by $I$ and the touched $1$s create the derangements. But isn't there some clear method to prove it?
I am new to group theory. Please ask for clarifications in case of any discrepancies. Any hint will be a great help!
 A: Here's a way to transform your great illustration into a proof:
Let $e_1,\dots,e_n$ be the standard basis of $\Bbb R^n$, then $P$ and $Q$ act on these exactly as the corresponding permutations.
So, by the condition, each $e_i$ is either moved by $P$ and fixed by $Q$, or is moved by $Q$ and fixed by $P$, moreover then $Qe_i$ is still in the cycle of $Q$, thus it's also fixed by $P$.
In the first case we have $Qe_i=e_i$,
$$PQe_i=Pe_i=Pe_i+e_i-e_i=(P+Q-I)e_i\,.$$
While in the second case we have $Pe_i=e_i$ and $PQe_i=Qe_i$,
$$PQe_i=Qe_i=e_i+Qe_i-e_i=(P+Q-I)e_i\,.$$
Since this holds for each element of a basis, it holds for all vectors, and hence $PQ=P+Q-I$.
(Note that this proof also allows $P$ and $Q$ to have more cycles, and the condition that we really need here is that the sets of fixed points of $P$ and $Q$ are disjoint.)
A: An alternative approach: note that
$$
PQ = P + Q - I \iff PQ - P - Q + I = 0 \iff (P - I)(Q - I) = 0.
$$
From there, note that if $P$ corresponds to a permutation of the elements of $M \subset \{1,\dots,n\}$ (i.e. all elements not in $M$ are fixed), then we will have $(P - I)e_i = 0$ for all $i \notin M$.
A: Not an answer but a generalisation of the above:-

Let $P_{1}, P_{2},\cdots P_{n}$ ($n\gt 2$) represent disjoint permutation cycles, we'll have $$\prod_{i=1}^{n}P_{i} = \sum_{i=1}^{n}P_{i}-(n-1)I$$

Proof :- We'll prove it by induction. It holds for $n=2$, as proved in the problem. Let it holds for $n = k$ i.e. $$\prod_{i=1}^{k}P_{i} = \sum_{i=1}^{k}P_{i}-(k-1)I$$ we'll prove it also holds for $n=k+1$. Let $P_{k+1}$ represent permutation cycle disjoint to $P_{1},P_{2}\cdots P_{k}$ then  $$\biggl(\prod_{i=1}^{k}P_{i}\biggr)P_{k+1} = P_{1}P_{k+1}+P_{2}P_{k+1}+\cdots P_{k}P_{k+1}-(k-1)P_{k+1}.$$ We also have $$P_{1}P_{k+1} = P_{1}+P_{k+1}-I $$ $$P_{2}P_{k+1} = P_{2}+P_{k+1}-I$$ $$ \vdots$$ $$P_{k}P_{k+1} = P_{k}+P_{k+1}-I$$ adding these equations we get $$P_{1}P_{k+1}+P_{2}P_{k+1}\cdots P_{k}P_{k+1} = \sum_{i=1}^{n}P_{i}+kP_{k+1}-kI$$ $$P_{1}P_{k+1}+P_{2}P_{k+1}\cdots P_{k}P_{k+1}-(k-1)P_{k+1} = \sum_{i=1}^{n}P_{i}+kP_{k+1}-kI-(k-1)P_{k+1}$$ $$P_{1}P_{k+1}+P_{2}P_{k+1}\cdots P_{k}P_{k+1}-(k-1)P_{k+1} = \sum_{i=1}^{k+1}P_{i}-kI$$  which gives $$\prod_{i=1}^{k+1}P_{i}=\sum_{i=1}^{k+1}P_{i}-kI$$
