- #1

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## 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.