# nontrivial example of permutations (discrete)

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?

• 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. – badjohn Apr 9 '17 at 14:25
• Isn't $\pi^{(3)} = (1,3,5)$? So it's not equal to the identity function – user384262 Apr 9 '17 at 14:30
• 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. – badjohn Apr 9 '17 at 14:48

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