Example: The permutation σ S 5 with σ(1) = 3, σ(2) = 5, σ(3) = 4, σ(4) = 1, σ(5) = 2 is written

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1 Notation: Throughout, let N n := {1,, n} Definition 01 Let n N A permutation of N n is a bijection N n N n We write S n for the set of permutations of N n The set S n together the function S n S n S n that maps α, β to the composition of functions α β is a group We call this group the symmetric group on n elements Why is S n a group? i If α, β S n then α β is bijective and thus α β S n ii The identity map ɛ : N n N n, defined by ɛi := i for all i N n, is the identity element for S n iii Bijective maps have inverses If α S n then there exists β S n such that α β = ɛ iv Multiplication is associative since function composition is always associative Notation: From now on, for α, β S n we will write αβ to mean α β For a permutation σ of N n, we write: 1 2 n σ1 σ2 σn Example: The permutation σ S 5 with σ1 = 3, σ2 = 5, σ3 = 4, σ4 = 1, σ5 = 2 is written Definition 02 If σ S n has the property that there exist a 1,, a m N n such that σa i = a i+1, for 1 i m 1; σa m = a 1, and σx = x, for x / {a 1,, a m } we say σ is an m-cycle and write σ in cycle notation as a 1 a 2 a m A transposition is a 2-cycle Example: The permutation σ := is a 3-cycle We write σ in cycle notation as 142 Definition 03 We say α, β S n are disjoint if, {x αx x} {x βx x} = 1

2 Example: Let and σ := τ := γ := The permutations σ and τ are disjoint but σ and γ are not disjoint Lemma 04 Let α 1,, α m S n be pairwise disjoint permutations and let τ S n The permutations α 1 α 2 α m and τ are disjoint if and only if α i and τ are disjoint for all 0 < i m Proof See exercise sheet Proposition 05 Every σ S n can be written as a product of disjoint cycles Proof Fix n N We shall prove the statement by induction on, Γσ := {a N n σa a} If Γσ = 0 then σ is the identity map on N n so σ = 12n Let σ S n Suppose k = Γσ > 0 and suppose the assertion is true for all permutations τ with Γτ < k Let i 0 N n be such that σi 0 i 0 Let i s := σ s i 0 Since N n is finite, there exists p, q N with p < q such that σ p i 0 = σ q i 0 Since σ is bijective, σ p q i 0 = i 0 Take r N least such that σ r+1 i 0 = i 0 Let τ be the r + 1-cycle, i 0 i 1 i r Now {a N n τ 1 σa = a} = {a N n σa = a} {i 0,, i r } So Γτ 1 σ < k = Γσ So, by the induction hypothesis, τ 1 σ can be written as a product of pairwise disjoint cycles, say τ 1 σ = α 1 α 2 α m So σ = τα 1 α 2 α m Since α 1 α 2 α m i j = τ 1 σi j = i j for 0 j m, the permutations α 1 α 2 α m and τ are disjoint By the lemma, this means τ and α i are disjoint for 0 < i m So σ is a product of disjoint cycles Example: The permutation

3 written as a product of disjoint cycles is Notation: Proposition 06 Every permutation on N n can be written as a product of transpositions Proof The identity is 1221 Since every permutation can be written as a product of cycles, it is enough to show that every cycle can be written as a product of transpositions Let i 1 i r S n be an r-cycle Then For i 1, For s > 1, i 1 i 2 i r = i 1 i r i 1 i r 1 i 1 i 3 i 1 i 2 i 1 i r i 1 i r 1 i 1 i 3 i 1 i 2 i 1 = i 1 i r i 1 i r 1 i 1 i 3 i 2 = i 2 i 1 i r i 1 i r 1 i 1 i 3 i 1 i 2 i s = i 1 i r i 1 i r 1 i 1 i s+1 i 1 i s i s = i 1 i r i 1 i r 1 i 1 i s+2 i 1 i s+1 i 1 = i 1 i r i 1 i r 1 i 1 i s+2 i s+1 = i s+1 Example: The permutation 123 S 4 can be written as both 1312 and So factorisation into transpositions is not unique, even more, the number of transpositions used in a factorisation is not unique So, what is unique? In order to answer this question we first need to define the action of a permutation σ S n on a function from Z n to Z Reminder Z n := Z } {{ Z } n times Let σ S n and f : Z n Z be a function We define σf to be the function from Z n Z defined by σfx 1,, x n := fx σ1,, x σn Example: Let f : Z 3 Z be the function defined by fx 1, x 2, x 3 := x 1 x 2 + x 3 and σ := 123 S 3 The function σfx 1, x 2, x 3 = fx 2, x 3, x 1 = x 2 x 3 + x 1

4 Lemma 07 Let σ, τ S n and f, g : Z n Z Then i στf = στf ii σfg = σfσg Proof See exercise sheet Theorem 08 There is a map sign : S n {1, 1} such that: a For every transposition τ S n, signτ = 1 b For permutations σ, σ signσσ = signσsignσ This function is unique with these properties signσ the signature of σ For σ S n, we call Proof Fix n N Let : Z n Z be the function defined by x 1,, x n := x j x i 1 i<j n Claim: For a transposition τ S n, τ = Let τ = rs with r < s By lemma 07i τ x 1,, x n = τx j x i 1 i<j n Clearly, if i, j / {r, s} then τx j x i = x j x i For the factor x s x r, we have that τx s x r = x r x s The remaining factors can be put into pairs as follows: x k x s x k x r, if k > s; x s x k x k x r, if r < k < s; x s x k x r x k, if k < r Each pair is unaffected by τ Therefore τ = So we have proved the claim Now suppose σ S n We can write σ = τ 1 τ m where τ 1,, τ m are transpositions By lemma 07ii, and by the claim So σ = or σ = σ = τ 1 τ 2 τ m τ 1 τ 2 τ m = 1 m For σ S n, let signσ = +1 if σ = and let signσ = 1 if σ = This map is well-defined since 1, 2,, n 0

5 Let σ, τ S n By lemma 07i, στ = στ So signστ = signσsignτ The function sign : S n {1, 1} is unique with properties a and b since every permutation is a product of transpositions Remark: Let σ S n and let τ 1,, τ m S n be transpositions such that σ = τ 1 τ m Then signσ = 1 m Definition 09 We call a permutation even if it can be written as a product of an even number of transpositions We call a permutation odd if it can be written as a product of an odd number of transpositions Corollary 010 A permutation σ is even if and only if signσ = 1 and is odd if and only if signσ = 1 Thus, a permutation can not be written as both a product of an even number transpositions and an odd number of transpositions