Prove $(\rho_1 \rho_2 \cdots \rho_r)^u = e \implies e=\rho_1^u=\rho_2^u=\cdots=\rho_r^u$

Question

In the proofwiki page: Order of Product of Disjoint Permutations

We have a product of disjoint permutations $\pi = \rho_1 \rho_2 \cdots \rho_r$. Why exactly is it that if $\pi^u=e$, then for each $\rho_s$ we have $\rho_s^u=e$, where $e$ is the identity permutation? I think we will have if they are not all $e$, then there will be at least 2 permutations $\rho_{s_1}$, $\rho_{s_2}$ where $\rho_{s_1}=\rho_{s_2}^{-1}$, which somehow contradicts the disjoint assumption, but I wasn't able to prove $\rho_{s_1}=\rho_{s_2}^{-1}$. I was able to prove only that $\rho_{s_1}^u=\rho_{s_2}^{-u}$, from which we can say

$\rho_{s_1}^u=\rho_{s_2}^{-u}$

$\implies \rho_{s_1}^u \rho_{s_2}^{u} = 1$

$\implies (\rho_{s_1} \rho_{s_2})^{u} = 1$

Answer 1

Fix some $k$ and let $S_k$ be the set which is acted upon by $\rho_k$ (the elements which $\rho_k$ actually moves). By disjointedness, none of the other $\rho_i$'s act on any of the elements of this set. So, when applying the entire permutation $u$ times, which results in the identity, the set $S_k$ only sees you apply $\rho_k^u$. Thus we must have $\rho_k^u=e$.

Answer 2

Note that if $\rho_i,\rho_j$ are disjoint permutations, then $\rho_i^n,\rho_j^m$ are also disjoint permutations for any $m,n\in\mathbb N$. Also note that if a product $\prod_i\rho_i$ of mutually disjoint permutations is $e$, then each of them is $e$, otherwise say $\rho_1=(1\ 2\cdots)$, then $\prod_i\rho_i$ sends $1$ to $2$, and is not $e$.

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Hope this helps.

Answer 3

$\rho_{s_1}$ and $\rho_{s_2}$ are disjoint if and only if $\rho_{s_1}^{m}$ and $\rho_{s_2}^{n}$ are disjoint.

Choose $m=n=u$.

If $(\rho_{s_1})^u (\rho_{s_2})^u = (\rho_{s_1} \rho_{s_2})^{u} = e$, then $\rho_{s_1}^{u}$ and $\rho_{s_1}^{u}$ are not disjoint because every permutation can be represented as the product of disjoint cycles. In this case, the representation $(\rho_{s_1})^u (\rho_{s_2})^u = e$ is not a disjoint representation of $e$. Therefore, $(\rho_{s_1})^u$ and $(\rho_{s_2})^u$ are not disjoint. Therefore, $\rho_{s_1}$ and $\rho_{s_2}$ are not disjoint. This contradicts our assumption.