Given a permutation $\sigma^n$ for certain $n$ in $N$, find the permutation $\sigma$
This is the proof, but i dont seem to understand it fully.
Given $\sigma^n$ for a certain $n$, how do we find $\sigma$ and when is $\sigma$ uniquely determined?
we will restrict ourselves to the situation of cyclic groups, such as the group $<\sigma>$, which is the group generated by the permutation $\sigma$, that is commutative.
Suppose we would know the order of $\sigma$ is $k$.
If gcd(n,k) = 1, then $n$ is not a mulitple of $k$, which means that $\sigma^n$ is not equal to the $Id$ permutation, since $\sigma^k = Id $ because we assumed that the order of $\sigma$ is $k$.
$$ \sigma^n \neq \sigma^{k*f} = (\sigma^{k})^{f} = Id^{f} = Id $$
Then, in this case, $\sigma$ is uniquely determined if $\sigma^n$ is given. (also, see comment by Blitzer)
$\sigma$ can then be calculated by computing the inverse $d$ where $ dn \equiv 1 \pmod{ k}$ or $ d \equiv n^{-1} \pmod{ k}$ since $\gcd(k,n) = 1$.(See source1 and source2.)
We can now construct the permutation $\sigma$ as follows: $$ (\sigma^n)^d = (\sigma^n)^{n^{-1}\pmod{k}} = \sigma^1. $$
If the $\gcd(k,n) = g > 1$, then we can write $n = g . n/g$. We can say,
$$\sigma^n = \sigma^{g . n/g}.$$
The order of $\sigma^{g}$ is equal to $\frac{k}{\gcd(k,g)}$. (See source0.)
Thus, |$\sigma^{g}$| = k/g since,
$$ gcd(k,g) = gcd(k,gcd(k,n))= gcd(gcd(k,k),n) = gcd(k,n) = g $$
We want to construct $\sigma^g$ from $\sigma^n$ as follows:
$$ (\sigma^n)^d = (\sigma^{g . n/g})^{(g/n)^{-1}\pmod{k/g}} = \sigma^g. $$
this is only possible if $gcd(n/g,k/g) = 1$ so that there exists a unique inverse $d$. (see source1 and source2.)
$$ gcd(n/g,k/g) = 1/g* gcd(n,k) = 1/g * g = 1$$
Now, we know the permutation $\sigma^g$ with $g|k$. Assume $\sigma^{'}$ a random solution, then
$$\sigma^g = \sigma^{'g} => (\sigma^g / \sigma^{'g}) = Id$$
This proves that a random solution is of the form $\sigma * \zeta_g$, with $\zeta_g$ a random element of the order which is a devider of $g$. Because we have restricted ourselves to the group generated by $\sigma$, the solutions are: $$ \sigma * \sigma^{ik/g}, \space for \space i = 0,...,g-1$$
To find a random solution, in the special case that $gcd(|\sigma^g|,g) = 1$, a solution of $\sigma^g$ is given by $(\sigma^g)^{g^{-1}(\mod |\sigma^g|)}$. If $gcd(|\sigma^g|,g) > 1$, then there doesn't exist a simple formula. (algorithms)
Questions:
- This whole time we assumed the order of $\sigma$ is $k$, but how do we know the order $\sigma$ and how do we know if $gcd(k,n)$ is 1 if we dont know $k$?
- from the moment "Now, we know the permutation $\sigma^g$ with $g|k$." i dont get it anymore. Maybe somebody is familiar with this proof and can explain the last bit?
To the first question, i have partially found how we would know the value k if gcd(k,n) = 1, but still, we would need to assume that the gcd is 1, which doesnt help. Maybe we need to work in cases each time we have a permutation \sigma^n ?
Here my Partial Answer:
If gcd(k,n) = 1, then the order of $\sigma^n$ is equal to $\frac{k}{\gcd(k,n)}$. (See source0.).
Then the orders are equal to each other, $|\sigma^n|$ = |$\sigma$|/1.
So, If we know the order of $\sigma^n$, we will know the order of $\sigma$.
In order to find the order of $\sigma^n$, we need to find the smallest integer $t$ such that $(\sigma^n)^t = Id $. We know $t$ becuase $\sigma^n$ is given.
Or, a simpler way of doing this, is by writing $\sigma^n$ as a composition of disjoint cycli. Where the order of $\sigma^n$ is equal to the lcm of the orders of the disjoint cycli. (see source3)
Sources:
Source0: How to prove $|a^k|=n/\gcd(n,k)$ whenever $|a|=n$?
Source1: Proving that modular inverse only exists when $\gcd(n,x)=1$
Source2: Uniqueness of modular multiplicative inverse
Source3: Why is the order of a $k$-cycle $\sigma$ equal to $k$?
