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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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    @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.