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Here is the exercise from a Pinter's "A book of abstract algebra" from a chapter dealing with permutations on a finite set:

Let $\alpha$ and $\beta$ be cycles, not neccessarily disjoint. Prove that, if $\alpha^2=\beta^2$, then $\alpha=\beta$.

I think I've found counterexample: $(2345)^2 = (24)(35) = (2543)^2$. Am I right?

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    Or even more simply, $(12)$ and $(1).$2011-12-17
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    Yes. The claim holds for cycles of odd length.2011-12-17
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    Thank you very much. Should I vote for deletion of this post now?2011-12-17
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    @furikuretsu: Not at all! Either jspecter or Jyrki can convert their comments to answers, or you can post your own answer to your question (this is in fact explicitly encouraged).2011-12-17
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    Go post an answer, @furikerutsu ! The Missus wants to go Christmas tree shopping, so I'm off air :-)2011-12-17
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    Sorry, I can't post answers on my own question untill 8 hours pass. So I am not an option too.2011-12-17
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    That's alright. Just make sure you post one in eight hours, okay?2011-12-17
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    Which book is it?2011-12-17
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    @lhf: It's Pinter book on algebra. I have added this info to question body.2011-12-17
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    Maybe you missed something earlier in the textbook. As Jyrki mentioned, if the length of the cycles is odd the exercies is easy to prove, while for each even length it is easy to find counterexamples. Check a little above, maybe there is a statement of the type " for the next examples the cycles have odd lenght" or something like that....2011-12-17
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    I've looked at the book (first and second edition) and it seems that the book is wrong.2011-12-17
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    @lhf, thank you very much for support.2012-02-11

1 Answers 1

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Yes, there really is mistake in a text of exercise.

The even simplier counterexample, submitted by jspecter: $(1)\not=(12)$, but $(1)^2=(12)^2$

For cycles of odd length, as Jyrki Lahtonen noted, the exercise claim holds.