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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 $(1).$ – jspecter Dec 17 '11 at 8:21
Yes. The claim holds for cycles of odd length. – Jyrki Lahtonen Dec 17 '11 at 8:22
@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). – Zev Chonoles Dec 17 '11 at 8:41
Go post an answer, @furikerutsu ! The Missus wants to go Christmas tree shopping, so I'm off air :-) – Jyrki Lahtonen Dec 17 '11 at 8:48
Which book is it? – lhf Dec 17 '11 at 13:17
up vote 5 down vote accepted

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.

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