# Powers of permutations

1. Nov 19, 2017

### fishturtle1

1. The problem statement, all variables and given/known data
In the following problems, let $\alpha$ be a cycle of length s, say $\alpha = (a_1a_2 ... a_s)$

3)Find the inverse of $\alpha$ and show that $\alpha^{-1} = \alpha^{s-1}$
2. Relevant equations
I've observed in the previous problem that there are $s$ distinct powers of $\alpha$, similar to how $\mathbb{Z}_n$ has n elements.

3. The attempt at a solution
The inverse of $\alpha$ is $\alpha^{-1} = (a_sa_{s-1} ... a_2a_1)$

I need to show $\alpha^1 = \alpha^{s-1}$
I've tried $\alpha^{-1}\alpha^1 = \alpha^{-1}\alpha^{s-1}$
=> $e = \alpha^{s-2}$

and $\alpha^1\alpha^{-1} = \alpha^{s-1}\alpha^{-1}$
=> $e = \alpha^{s-2}$

and $\alpha^{-1}\alpha^s = \alpha^{s-1}\alpha^s$
=> $\alpha{s-1} = \alpha^{2s-1}$, and since there are only $s$ distinct permutations, maybe somehow we can conclude $\alpha^{2s-1} = \alpha^{2s-1-s} = \alpha^{s-1}$.. But if that's the case, I can use that same argument starting with $\alpha^1 = \alpha^{s-1}$

I am asking for a small hint.

2. Nov 19, 2017

### haruspex

3. Nov 19, 2017

### StoneTemplePython

hint: raising to what power leaves things intact? i.e. what $\alpha$ to the what power is the identity operation?
- - - -
note: it's perhaps a matter of taste (and what course you are in), but I would use permutation matrices to show this. Then again, I really like matrices and matrix groups.

4. Nov 19, 2017

### fishturtle1

OK i made a mistake reading the question, sorry about that, but think I can use post 3's hint..

We observed that raising a permutation, $\alpha$ of length s, to the $s^{th}$ power, we have $\alpha^s = e$.
Therefore $\alpha^{-1} = \alpha^{-1}e = \alpha^{-1}\alpha^s = \alpha^{s-1}$ []