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I was given these two permutations:

$\pi = (1,3,5)(2,4)(7,9)$ and $\sigma = (5,4,8,9)$.

I had to write out $\pi \circ \sigma$ in cyclic notation which is just $(1,3,5,2,4,8,7,9)(6)$, I think.

But then I was asked to "give a specific example of a nontrivial (i.e. not the identity) permutation $\pi \in S_{n}$ such that $\pi^{(2)}=\pi^{-1}$."

But I have no idea what it means for a permutation to be nontrivial. I wrote out $\pi^{(2)}$ which is $(1,5,3)(2)(4)(6)(7)(8)(9)$ and $\pi^{-1}$ which is $(1,5,3)(2,4)(6)(7,9)(8)$. So they're not equal, so I'm not sure how to answer the question.

I looked online and I think when it says that it has to be nontrivial it means that no number can map to itself?

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  • $\begingroup$ You say that you have no idea what non-trivial means but you just said it yourself: it is the do nothing identity permutation. If it were not for this restriction then the identity would be an answer since $\pi^2$ and $\pi^{-1}$ would both also be the identity and hence the same. You want $\pi^2$ and $\pi^{-1}$ to be the same. Think of what $\pi^3$ must be. $\endgroup$ – badjohn Apr 9 '17 at 14:25
  • $\begingroup$ Isn't $\pi^{(3)} = (1,3,5)$? So it's not equal to the identity function $\endgroup$ – user384262 Apr 9 '17 at 14:30
  • $\begingroup$ I read it to mean: find another permutation that has the specified property. Indeed, the $\pi$ quoted here does not have that property and $\sigma$ does not either. As Marc comments, "non-trivial" is not a standard term but I would interpret it as just "non-identity" not that it must permute all of the elements. So, I would say that $(1, 2, 3)$ is not trivial. $\endgroup$ – badjohn Apr 9 '17 at 14:48
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You found $\pi \circ \sigma$, $\pi^{(2)}$ and $\pi^{-1}$ correctly. And yes, $\pi^{(2)} \neq \pi^{-1}$

What you are probably asked to do is to come up with any different non-trivial permutation $\pi \neq \text{id}$ for which this equality holds.

As an example, you may take $\pi = (1 2 3)(4)$, then $\pi^3 = \text{id}$ which means $\pi^{(2)} = \pi^{-1}$.

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Hint: $\pi^{2}=\pi^{-1}$ iff $\pi^{3}=id$.

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Your question already answers the question what "nontrivial" means for a permutation, since the use of "nontrivial" is immediately followed by the parenthesised remark "(i.e. not the identity)". So in this context a nontrivial permutation is any permutation except the identity. This use of "nontrivial" is by the way not standard use ("non identity" is clearer).

So the question asks for a permutation that moves at least one element (thus avoiding to be the identity), but which when composed with itself gives the identity (moves nothing at all). Clearly any element that does move under the permutation must move back to its original position under the second application of that permutation.

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