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Claim:

Let $\sigma \in S_n$, be a permutation, and $X = \{1, 2, ..., n\}$. The cycle decomposition of $\sigma$ can be recovered by considering the orbits of the action of $\langle\sigma\rangle$ on $X$.

Let $\langle \sigma \rangle := G$. By the Orbit-Stabilizer Theorem, the action of $G$ partitions $X$ into unique orbits $O(X)$, and there exists a bijection $f : g \cdot \mathrm{Stab}(X) \mapsto g \cdot X$.

$G = \{e, \sigma, \sigma^2, ..., \sigma^{k-1}\}$ for $G$ of order $k$.

$\mathrm{Stab(X)} = \{\sigma^i \in G : \sigma^i(X) = X\} = O(X)$

Thus the bijection maps $\sigma^i(X) \mapsto \sigma^i(O(X))$ $G$ is cyclic, therefore $G/O(G)$ is cyclic and by Lagrange's theorem we get that $|O(X)| = |G : O(X)| = k$

Thus the cosets of $O(X)$ in $G$ correspond to the permutation $\{O(X), \sigma(O(X)), ..., \sigma^{k-1}(O(X))\}$ where, by the bijection, each of these terms corresponds to $\{X, \sigma(X), ..., \sigma^{k-1}(X)\}$ respectively. Therefore, an orbit $|O(X)| = k$ corresponds to some $k$-cycle, and we get a cycle decomposition.

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  • $\begingroup$ Are you looking for a proof verification? $\endgroup$ – Mauve Apr 6 '18 at 17:50
  • $\begingroup$ I could swear I saw a very similar question earlier today (without solution), but I can't find it... Was that you? If yes, did you delete the question to post a new one? $\endgroup$ – Arnaud D. Apr 6 '18 at 18:01
  • $\begingroup$ Yes and yes. It was me, and I am looking for a verification, or to point out any mistakes in the proof, please. $\endgroup$ – The Bosco Apr 6 '18 at 18:03
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    $\begingroup$ Then you could have simply edited your first question with your attempt. But it's not a big deal ;) $\endgroup$ – Arnaud D. Apr 6 '18 at 18:08
  • $\begingroup$ Remark that $G/O(G)$ (it is a typo I think you mean $G/O(X)$) has no reason to be a group. I also don't understand why $X$ is $\{1,\dots,n\}$ at the beginning and seems to be an element of $X$ at the end (because as it is defined for the moment $Stab(X)=G$). To prove the claim, I think you are forced to write down explicitly the cycles arising in the decomposition of $\sigma$. My advice would be to work with $\sigma$ a cycle (I think I have already given the advice) and maybe work on some explicit example, for instance $(1,2)(4,5,6)$ in $S_6$. $\endgroup$ – Clément Guérin Apr 7 '18 at 3:09
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Your claim is obviously true: the cycle decomposition of$~\sigma$ is the collection of orbits of$~\sigma$ on$ ~X$. Or more precisely, that is the combinatorial cycle decomposition, which allows for cycles of length$~1$; a separate one for each fixed point. For the group theoretic cycle decomposition, there is a cyclic permutation for each combinatorial cycle of length${}>1$, obtained from$~\sigma$ by acting as$~\sigma$ on that cycle and as the identity on its complement (and $\sigma$ is then the product of those commuting cyclic permutations).

However your proposed "proof" is completely confused. The orbit-stabiliser theorem needs a base point $x_0\in X$, which you do not choose. The stabiliser then is $\operatorname{Stab}_G(x_0)=\{\,g\in G\mid g\cdot x_0=x_0\}\,$; your $\operatorname{Stab}(X)$ is meaningless (since every $g\in G$ globally maps $X$ to itself, the only meaning one could ascribe to that notation would be all of $G$, but that makes the notation pointless). When you say "bijection" you forget to mention the sets between which the bijection runs; for the orbit-stabiliser theorem it would be between $G/\operatorname{Stab}_G(x_0)$ and the orbit $G\cdot x_0$, but that does not correspond to what you write. So you are off to a very bad start, and there is no sense to be made of the remainder either.

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