Powers of permutations

  • #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.
 

Answers and Replies

  • #2
haruspex
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Homework Statement



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

I need to show ##\alpha^1 = \alpha^{s-1}##
Um... No. Reread the question.
 
  • #3
StoneTemplePython
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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
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Um... No. Reread the question.
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.
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}## []
 

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