Let $P$ is a permutation matrix which represents the permutation $\sigma\in S_{n}$. Let $\sigma_{1}$,$\sigma_{2}$,…$\sigma_{k}$ denote the disjoint permutations in the cycle form of $\sigma$.
Let $P_{i}$ and $c_{i}$ represents the permutation matrix corresponding to the permutations $\sigma_{i}$ and the cycle lengths of $\sigma_{i}$ respectively. Prove that P satisfies the equation (its characteristic polynomial)
$$\prod_{i=1}^{k}(P^{c_{i}}-I) =0$$
I know the following facts:
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For disjoint permutation matrices $P_{i},P_{j}$ we have $(P_{j}-I)(P_{i}-I) = 0$.
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For disjoint permutation matrices $P_{1},P_{2},\cdots,P_{k}$ we have $$\prod_{i=1}^{k}P_{i} = \sum_{i=1}^{k}P_{i}-(k-1)I.$$
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If P and Q are disjoint permutation matrices so are $P^{m}$ and $Q^{n}$ , $\forall m,n\in \Bbb N$.
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If P and Q are disjoint permutation matrices they commute.
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If P is a single-cycled permutation matrix with cycle length $k$ then $P^{k}=I$.
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Combining the fact $2$ and $3$ for disjoint permutation matrices $P_{1},P_{2},\cdots,P_{k}$ we also have $$\prod_{i=1}^{k}P_{i}^{m} = \sum_{i=1}^{k}P_{i}^{m}-(k-1)I, \forall m\in \Bbb N.$$
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Combining the fact $1$ and $3$ we have $(P_{j}^{n}-I)(P_{i}^{m}-I) = 0$ for any $n,m \in \Bbb N.$
MY TRY: I tried with an case where $P$ breaks in two single-cycled disjoint permutations $Q$ and $R$ with cycle lengths $m,n$ respectively. We need to prove that $$(P^{n}-I)(P^{m}-I) = \Bigr((QR)^{n}-I\Bigl)\Bigr((QR)^{m}-I\Bigl) = 0$$
Using fact $4$ and $5$
$$\Bigr((QR)^{n}-I\Bigl)\Bigr((QR)^{m}-I\Bigl)=\Bigr(Q^{n}R^{n}-I\Bigl)\Bigr(Q^{m}R^{m}-I\Bigl) = \Bigr(R^{n}-I\Bigl)\Bigr(Q^{m}-I\Bigl)$$ The fact $7$ as stated above states that it vanishes.
But it gets more calculative when P breaks in $3$ disjoint single cycled permutations. Also, generalisation would need more calculations.
I do not know Cayley hamilton theorem. I am new to group theory. Please ask for clarifications if something is not clear. Any hint would be a great help.
Best Answer
I think this is much easier than you are making it out to be. By relabelling, you may assume that the permutation is $$(1,\dots,c_1)(c_1+1,\dots,c_1+c_2)...$$ The permutation matrix of this product of disjoint cycles is a block-diagonal matrix, with the blocks being the permutation matrices of each cycle.
Products and sums of block-diagonal matrices are block-diagonal, obtained by taking the products and sums of each block. Thus a block-diagonal matrix satisfies a polynomial if and only if each block of it does.
Block $i$ certainly satisfies the polynomial $P^{c_i}-I$, your fact 5. Thus this is the zero matrix, and the product of this with anything else is zero, in particular your polynomial is the zero matrix on the $i$th block. Thus your matrix is zero.