Order of Cycle is Length of Cycle
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Theorem
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi \in S_n$ be a cyclic permutation of length $k$.
Then:
- $\order \pi = k$
where:
- $\order \pi$ denotes the order of $\pi$ in $S_n$.
Proof
Let $\pi = \tuple {a_0, a_1, \ldots, a_{k - 1} }$.
Observe that:
\(\ds \paren {\paren {j + n} \pmod k} + 1\) | \(=\) | \(\ds \paren {j + n + 1} \pmod k\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \pi {a_{\paren {j + n} \pmod k} }\) | \(=\) | \(\ds a_{\paren {j + n + 1} \pmod k}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall n \in \N, \forall j \in \Z / k \Z: \, \) | \(\ds \map {\pi^n} {a_j}\) | \(=\) | \(\ds a_{\paren {j + n} \pmod k}\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall j < k: \, \) | \(\ds \pi^k\) | \(=\) | \(\ds I_{S_n}\) | ||||||||||
\(\, \ds \land \, \) | \(\ds \pi^j\) | \(\ne\) | \(\ds I_{S_n}\) | where $I_{S_n}$ denotes the identity mapping on $S_n$ |
Hence the result, by definition of order of group element.
$\blacksquare$
Sources
- Jonathan Y. (https://math.stackexchange.com/users/89121/jonathan-y), Why is the order of a k-cycle σ equal to k?, URL (version: 2014-02-12): https://math.stackexchange.com/q/673710