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On page 66 of these notes is proposition 4.26:

Every permutation can be written (in essentially one way) as a product of disjoint cycles.

The proof begins as follows:

Let $\sigma \in S_n$, and let $O \subseteq \{1,...,n\}$ be an orbit for $\langle \sigma \rangle$....

What does it mean for $O$ to be an orbit for $\langle \sigma \rangle$? I am unfamiliar with this terminology. From what I gather, the implicit action is of $S_n$ on $\{1,...,n\}$ by functional evaluation. So, $O$ will be the orbit of some element in $\{1,...,n\}$. How can it be an orbit for $\langle \sigma \rangle$?

EDIT

Also, the author writes $O = \{i,\sigma (i),..., \sigma^{r-1}(i)\}$. How do we know this equality holds? What if $\sigma$ has order smaller than $r-1$?

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    $\begingroup$ How did they define $r$? $\endgroup$ – Mike Earnest Mar 14 '19 at 15:32
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    $\begingroup$ $O$ is the orbit of $i$ under the action of the cyclic subgroup $\langle \sigma \rangle$ [this action naturally being the restriction of the action of $S_n$]. $\endgroup$ – M. Vinay Mar 14 '19 at 15:34
  • $\begingroup$ See also Exercise 7 on UMN Fall 2017 Math 4990 homework set #7 for a version of this proof with all details filled in. It is one of the most painful to formalize proofs in basic abstract algebra. (Note that my $\sim$-equivalence classes are exactly the orbits of $\sigma$, although I define them a bit differently.) $\endgroup$ – darij grinberg Mar 14 '19 at 15:37
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Call $I_n:=\{1,\dots,n\}$. For any given $\sigma \in S_n$, consider the map $\mathcal{A}_\sigma\colon \langle\sigma\rangle \times I_n \rightarrow I_n$, $(\sigma^k,i) \mapsto \sigma^k(i)$. Now:

  1. for $k=o(\sigma)$, $\iota_{S_n}=\sigma^{o(\sigma)} \in \langle \sigma \rangle$ and $(\iota_{S_n},i)=\iota_{S_n}(i)=i, \forall i \in I_n$;
  2. $(\sigma^k\sigma^l,i)=(\sigma^k\sigma^l)(i)=\sigma^k(\sigma^l(i))=(\sigma^k,(\sigma^l,i))$

so that $\mathcal{A}_\sigma$ is an action. In particular, the orbit "by the point" $j \in I_n$ is the set:

$$O_\sigma(j):=\{\sigma(j),\dots,\sigma^{l_j}(j)=j\} \tag 1$$

where $l_j$ $(\le o(\sigma))$ is the least value of the exponent $k$ such that $\sigma^k(j)=j$.

The number of orbits is given by:

$$r=\frac{1}{o(\sigma)}\sum_{k=1}^{o(\sigma)}\operatorname{Fix}(\sigma^k)=\frac{1}{o(\sigma)}\sum_{j=1}^{n}\operatorname{Stab}(j) \tag 2$$

where $\operatorname{Fix}(\sigma^k):=\{j \in I_n\mid \sigma^k(j)=j\}$ and $\operatorname{Stab}(j):=\{\sigma^k \in \langle\sigma\rangle\mid \sigma^k(j)=j\} \le \langle\sigma\rangle$. Therefore, $\exists \{j_1,\dots,j_r\} \subseteq I_n$ such that:

$$I_n = \bigsqcup_{k=1}^{r}O_\sigma(j_k) \tag 3$$

Note that $\sigma_{|O_\sigma(j_k)}$ is a bijection on $O_\sigma(j_k)$, so that its extension to $I_n$ by the identity map, say $\alpha_k$, is a bijection on $I_n$, namely $\alpha_k \in S_n$. One could prove that $\alpha_k^{l_k}=\iota_{S_n}$, where $l_k$ is defined above (so, $\alpha_k$ is a $l_k$-cycle), and that $\sigma=\alpha_1\dots\alpha_r$: this is the decomposition of a permutation into disjoint cycles, which depends on $\sigma$ solely. "Disjoint" because $supp(\alpha_k)=O_\sigma(j_k)$. This is addressed in the Edit of this post of mine.

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