# Powers of permutations

## Homework Statement

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}##

## Homework 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.

## 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.

haruspex
Homework Helper
Gold Member
2020 Award

## Homework Statement

3)... show that ##\alpha^{-1} = \alpha^{s-1}##

I need to show ##\alpha^1 = \alpha^{s-1}##

StoneTemplePython
Gold Member
hint: raising to what power leaves things intact? i.e. what ##\alpha## to the what power is the identity operation?
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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.