I'm reading the proof of Theorem 14 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.

Any permutation $\sigma \neq 1$ on a finite set $X$ is a composite $\gamma_{1} \cdot \cdot \gamma_{k}$ of disioint cyclic permutations $\gamma_{i}$, each of length $2$ or more. Except for changes in the order of the cyclic factors, $\sigma$ has only one such decomposition.

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They said that

Each of the points ${\sigma}^{i} {x}$ in this set $C$ has the same orbit [under $\sigma$].

IMHO, this is only when ${\sigma}^{i} {x}$ is a generator of $C$, i.e. $\gcd(i,m) = 1$. As such, I feel that the statement may be not correct.

Could you please verify my observation?

  • 1
    $\begingroup$ He means that the orbit is under the permutation $\sigma$, not under the permutation $\sigma^i$ $\endgroup$ – fhn Jul 1 '20 at 15:55

If $y=\sigma^i(x)$, $x=\sigma^{-i}(y)$, we have $\sigma^p=Id$ implies that $\sigma^{-i}=\sigma^{p-i}$ and $x=\sigma^{p-i}(y)$.


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