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Conjugating a cycle relabels each entry: $g (a_{1} \dots a_{k}) g^{- 1} = (g (a_{1}) \dots g (a_{k}))$

Statement

For $g \in S_{n}$ and a cycle $c = (a_{1} a_{2} \dots a_{k})$ , $g c g^{- 1} = (g (a_{1}) g (a_{2}) \dots g (a_{k})) .$ For a $1$ -cycle this says that a fixed point $a_{1}$ is relabelled as the fixed point $g (a_{1})$ .

Facts & Assumptions

Given: A permutation $g \in S_{n}$ and a cycle $c = (a_{1} \dots a_{k})$ .

[F1]

Permutation products act with the right factor first, and cycle notation records the successive images of the displayed entries (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition).

[F2]

The support of a cycle is its set of moved entries; entries outside it are fixed (Support, fixed points, disjoint cycles, cycle length, disjoint-cycle decompositions, and cycle type).

Proof

technique · direct

1.1

For each $j$ modulo $k$ , $(g c g^{- 1}) (g (a_{j})) = g (c (a_{j})) = g (a_{j + 1})$ .

1.2

If $y$ is outside ${g (a_{1}), \dots, g (a_{k})}$ , then $g^{- 1} (y)$ is outside the support of $c$ , so [F2] gives $(g c g^{- 1}) (y) = y$ .

F2algebra

2.1

Steps 1.1--1.2 describe exactly the cycle on the right, including the $k = 1$ case.

F1step 1.1step 1.2∎

Depends on

Used by

Z(Sₙ) is trivial for n≥3 Corollary
Conjugating (1 4)(2 5 3) by an explicit permutation in S₅ Example
V₄={1,(12)(34),(13)(24),(14)(23)} is a proper nontrivial normal subgroup of A₄ Example
Every subnormal series is a normal series False statement
FALSE: Aₙ is simple for every n≥4 False statement
FALSE: two even permutations of the same cycle type are always conjugate in Aₙ False statement
K is normal in HcharG always implies K is normal in G False statement
Every nontrivial normal subgroup of Aₙ contains a 3-cycle for n≥5 Lemma
The symmetric groups Sₙ are solvable for n≤ 4 Lemma
For n≥2, (1 2 … n) and (1 2) generate Sₙ Theorem
For prime p, a transitive subgroup of Sₚ containing a transposition is all of Sₚ Theorem
Two elements of Sₙ are conjugate if and only if they have the same cycle type Theorem

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Conjugacy Classes (standard reference, not scraped)

Conjugating a cycle relabels each entry: g(a1 … ak)g−1=(g(a1) … g(ak))g(a1​…ak​)g−1=(g(a1​)…g(ak​))