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Hi all,

Let $p=2^n+1$ be an arbitrary (Fermat) prime. Consider $\pi=(1,2,...,p)\in S_p$, the symmetric group of degree $p$. My question is:

Is there always a cycle, denoted by $\rho$, of length $p-1$, such that $$\rho \pi \rho^{-1}=\pi^2.$$

I know the answer is yes for $p=3,5$ but not sure if it is true in general. I think the question should be easy to answer but my head is blank now...

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    $\begingroup$ Since you seem to be creating new accounts spontaneously, I suggest you register before doing anything else. $\endgroup$
    – S. Carnahan ♦
    Commented Apr 7, 2013 at 9:51

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Not when $p >5$. Conjugation by $\rho$ induces an automorphism of order $2n$ of $\langle \pi \rangle.$ Since $\pi$ commutes with nothing other than its powers, this means that $\rho$ must be an element of order $2n.$ This is $p-1$ when $p =3$ or $p =5,$ but not for bigger Fermat primes.

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