Order of a permutation is the lcm of the orders of cycles

[Mr Davis 97]

Let $\sigma \in S_n$, and $|\sigma| = r$. Then, we'll assume that $\sigma$ can be decomposed into a product of disjoint cycles, which commute with each other. So $\sigma = c_1c_2 \dots c_k$. Then $\sigma^r = (c_1c_2 \dots c_k)^r$ = 1. Since the cycles commute, we have $c_1^rc_2^r \dots c_k^r = 1$. Now, I am a little stumped. How can I conclude that $c_1^r = c_2^r = \cdots = c_k^r = 1$?

 

[fresh_42]

Let $\sigma \in S_n$, and $|\sigma| = r$. Then, we'll assume that $\sigma$ can be decomposed into a product of disjoint cycles, which commute with each other. So $\sigma = c_1c_2 \dots c_k$. Then $\sigma^r = (c_1c_2 \dots c_k)^r$ = 1. Since the cycles commute, we have $c_1^rc_2^r \dots c_k^r = 1$. Now, I am a little stumped. How can I conclude that $c_1^r = c_2^r = \cdots = c_k^r = 1$?

Let's say we have the following order: $n = \sigma^r (n)= (c_1c_2 \dots c_k)^r(n)= (c_1^r(c_2^r( \dots (c_k^r)(n)\ldots )$. Now assume $n$ is one of the numbers, which are transformed by $c_k$. What can you say about $c_k^r(n)$ and why?

 

[Mr Davis 97]

Let's say we have the following order: $n = \sigma^r (n)= (c_1c_2 \dots c_k)^r(n)= (c_1^r(c_2^r( \dots (c_k^r)(n)\ldots )$. Now assume $n$ is one of the numbers, which are transformed by $c_k$. What can you say about $c_k^r(n)$ and why?

Well either $c_k^r = 1$ or $c_k^r = (c_1^rc_2^r \dots c_{k-1}^r)^{-1} \ne 1$, but the latter is not possible because the cycles, their powers, and their inverses are all disjoint?

 

[fresh_42]

Concentrate on $n$, a number which is transformed by $c_k$. What do the consecutive $c_i\, , \,i<k$ to $c_k(n)$ resp. $c_k^r(n)\,?$ Can they reverse the transformation done by $c_k\,?$

 

[Mr Davis 97]

Concentrate on $n$, a number which is transformed by $c_k$. What do the consecutive $c_i\, , \,i<k$ to $c_k(n)$ resp. $c_k^r(n)\,?$ Can they reverse the transformation done by $c_k\,?$

Well... If $C_k^r = (c_1^rc_2^r \dots c_{k-1}^r)$ and $c_k^r$ are disjoint, then if $n$ is transformed by $c_k^r$ then $C_k^r$ can't possibly transform $c_k^r(n)$ back to $n$ (since we know $C^r_k c^r_k = 1$, $C^r_k$ would have to). So $c_k^r = 1$.

 

[fresh_42]

Well... If $C_k^r = (c_1^rc_2^r \dots c_{k-1}^r)$ and $c_k^r$ are disjoint, then if $n$ is transformed by $c_k^r$ then $C_k^r$ can't possibly transform $c_k^r(n)$ back to $n$ (since we know $C^r_k c^r_k = 1$, $C^r_k$ would have to). So $c_k^r = 1$.

You're thinking far too complicated. Say we have two disjoint permutations $\sigma , \tau$. Now assume that $\tau(n) = m$. So we have $\sigma (\tau (n)) = \sigma (m)$. Do we already know the result of $(\sigma \tau)(n)\,?$