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

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

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

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

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

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 $$\rho_k^u=e$$.

• "By disjointedness, none of the other $\rho_i$'s act on any of the elements of this set." This intuition is EPIC! Thank you! – user198044 Oct 10 '18 at 7:08
• @JackBauer What did you think "disjoint" meant? – Arthur Oct 10 '18 at 7:10
• Just that they commute (I didn't realize other properties were relevant, so I didn't bother to analyze the definition of disjoint), which I'm now realizing is that weaker than disjoint. :) – user198044 Oct 10 '18 at 7:11
• Arthur, may I please double check: inverses are not disjoint, is that right? – user198044 Oct 10 '18 at 7:15
• @JackBauer Apart from the identity permutation (which is its own inverse and still disjoint), no they're not. – Arthur Oct 10 '18 at 7:18

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 $$e$$.

Hope this helps.

• I think your answer, at least for the second sentence, argues against exactly one of them is not $e$. My argument is that if one of them is not $e$, then another one isn't $e$ too. To have the product equal to $e$, one is the inverse of the other so $\rho_{s_1}^u=\rho_{s_2}^{-u}$. As for your first sentence, what does it mean for $\rho_{s_1}^u$ and $\rho_{s_2}^{u}$? Inverses are not disjoint? – user198044 Oct 10 '18 at 7:11
• I was trying to express the same idea as that of Arthur in fact. Because other permutations will not affect the elements acted by any one of the permutations, if the product is $e$, then everyone of them is $e$. – awllower Oct 10 '18 at 9:09

$$\rho_{s_1}$$ 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.