10 Chapter 1. Groups I Cycle Structure Number (1) 1 (1 2) 6 (1 2 3) 8 (1 2 3 4) 6 (1 2)(3 4) 3 24 Table 1. Permutations in S4. Cycle Structure Number (1) 1 (1 2) 10 (1 2 3) 20 (1 2 3 4) 30 (1 2 3 4 5) 24 (1 2)(3 4 5) 20 (1 2)(3 4) 15 120 Table 2. Permutations in S5. then αγα−1 is the permutation obtained from γ by applying α to the symbols in the cycles of γ. Remark. For example, if γ = (1 3)(2 4 7)(5)(6) and α = (2 5 6)(1 4 3), then αγα−1 = (α1 α3)(α2 α4 α7)(α5)(α6) = (4 1)(5 3 7)(6)(2). Proof. Observe that αγα−1 : α(i1) → i1 → i2 → α(i2). (1) Let σ denote the permutation defined in the statement. If γ fixes i, then σ fixes α(i), for the definition of σ says that α(i) lives in a 1-cycle in the factorization of σ. Assume that γ moves a symbol i say, γ(i) = j, so that one of the cycles in the complete factorization of γ is (i j . . . ). By definition, one of the cycles in the complete factorization of σ is ( α(i) α(j) . . . ) that is, σ : α(i) → α(j). Now Equation (1) says that αγα−1 : α(i) → α(j), so that σ and αγα−1 agree on all numbers of the form α(i). But every k ∈ X = {1,...,n} lies in im α, because the permutation α is surjective, and so σ = αγα−1. • Example 1.8. We illustrate the converse of Lemma 1.7 the next theorem will prove that this converse holds in general. In S5, place the complete factorization of a 3-cycle β over that of a 3-cycle γ, and define α to be the downward function. For example, if β = (1 2 3)(4)(5) γ = (5 2 4)(1)(3), then α = 1 2 3 4 5 5 2 4 1 3 , and the algorithm gives α = (1 5 3 4). Now α ∈ S5 and γ = (α1 α2 α3),

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