Can the matrix representation of a product of disjoint cycles be made up of the matrix representation of each cycle?

Question

Let $\rho \in S_n$ be a permutation and its matrix representation be $M(\rho)$ where $M: S_n \rightarrow GL(\mathbb{C}^n)$.

Then writing out $M(\rho)$ is very easy. But now if we represent $\rho$ as the product of disjoint cycles, ie $\rho = (\rho_1 \rho_2 \cdots \rho_n)$ where each $\rho_i$ is a cycle, then is there a way of writing the matrix representation of this product by using the matrix representation of each cycle. i.e. can we write $M(\rho)$ by using each $M(\rho_i)$.

For example if we have $(1 2 3)(4 5)$, its matrix representation is $ \begin{bmatrix} 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{bmatrix} $ which is in block diagonal form (I don't know how to latex that) and so clearly makes use of the matrix for both $(1 2 3)$ and $(4 5)$.

But if we have $(2 3 4)(1 5) \in S_5$, its matrix representation is $ \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix} $ which is not made up of the matrix for $(2 3 4)$ and the one for $(1 5)$.

The reason for this question is I am trying to find a formula for the determinant of a permutation minus the identity (i.e. $\det(M(\rho) - I)$). I believe that this is the product of each $\det(M(\rho_1) - I)$ but this would only make sense if $M(\rho)$ could be written as the block diagonal with each $M(\rho_i)$ being a block.

As a side note, why is the matrix representation of $(1 2 3)(4 5)$ not the matrix representation of $(1 2 3)$ times the matrix representation of $(4 5)$. The product of these two $5 \times 5$ matrices gives me $0$ instead of the result above. Intuitively should the matrix representation of the composition of two cycles not be the product of their matrix representations?

Thanks

Answer 1

If your intention is to compute $\det(M(\rho) - I)$,

then for $\rho=(a_1\,a_2\,\ldots\,a_k)(b_1\,b_2\ldots\,b_j)\ldots$

you might want to define a new basis $w_1,w_2,\ldots$ such that

$w_1=e_{a_1}$, $w_2=e_{a_2}$, ... , $w_k=e_{a_k}$,

$w_{k+1}=e_{b_1}$, $w_{k+2}=e_{b_2}$, ... , $w_{k+j}=e_{b_j}$, etc.

Denoting by $S$ the matrix that changes the vector basis from $\{e_i\}$ to $\{w_i\}$, we have that $SM(\rho)S^{-1}$ is in block form.

Now, $\det(M(\rho)-I)=\det(SM(\rho)S^{-1}-I)$.

$\\$

As for your side note,

As a side note, why is the matrix representation of (123)(45) not the matrix representation of (123) times the matrix representation of (45). The product of these two 5×5 matrices gives me 0 instead of the result above.

Have you made sure that the matrix representation of $(123)$ has diagonal entries $a_{4,4}=a_{5,5}=1$ instead of $0$? You can regard $(123)$ as $(123)(4)(5)$.