Conjugacy classes of a symmetric group are determined by cycle structure

Conjugate of an $𝑛$ -cycle is an $𝑛$ -cycle

Let $𝛼, 𝜏 \in 𝑆 𝑛$ where $𝛼$ an $𝑘$ -cycle with the form

𝛼 = (𝑎 1 𝑎 2 \dots 𝑎 𝑘 - 1 𝑎 𝑘)

where $𝑎 𝑖 \in ℕ 𝑛$ then the conjugate $𝜏 𝛼 𝜏 - 1$ is given by

𝜏 𝛼 𝜏 - 1 = (𝜏 (𝑎 1) 𝜏 (𝑎 2) \dots 𝜏 (𝑎 𝑘 - 1) 𝜏 (𝑎 𝑘))

and is hence also a $𝑘$ -cycle #m/thm/group/sym

Proof

Let $1 \leq 𝑖 \leq 𝑘$ . Then $𝜏 𝛼 𝜏 - 1 𝜏 (𝑎 𝑖) = 𝜏 𝛼 (𝑎 𝑖) = 𝜏 (𝑎 𝑖 + 1 m o d 𝑘)$ . For any $𝑏 \in ℕ 𝑛 ∖ {𝑎 𝑖} 𝑘 𝑖 = 1$ , $𝛼 (𝑏) = 𝑏$ , so $𝜏 𝛼 𝜏 - 1 𝜏 (𝑏) = 𝜏 (𝑏)$ . Hence $𝜏 𝛼 𝜏 - 1$ maps numbers of the form $𝜏 (𝑎 𝑖)$ to $𝜏 (𝑎 𝑖 + 1 m o d 𝑘)$ , and leaves all others invariant. Thus

𝜏 𝛼 𝜏 - 1 = (𝜏 (𝑎 2) 𝜏 (𝑎 3) \dots 𝜏 (𝑎 𝑘) 𝜏 (𝑎 1)) = (𝜏 (𝑎 1) 𝜏 (𝑎 2) \dots 𝜏 (𝑎 𝑘 - 1) 𝜏 (𝑎 𝑘))

as claimed.

This is a lemma for Conjugacy classes of a symmetric group are determined by cycle structure.

#state/tidy | #lang/en | #SemBr

Conjugate of an 𝑛-cycle is an 𝑛-cycle

Conjugate of an $𝑛$ -cycle is an $𝑛$ -cycle