6.2 Permutations continued

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Lauren Stokes
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1 6.2 Permutations continued Theorem A permutation on a finite set A is either a cycle or can be expressed as a product (composition of disjoint cycles. Proof is by (strong induction on the number, r, of points moved by a permutation. If r = 0 then ρ is the identity which is a 1-cycle. Assume result true for all r k. Let ρ be a permutation that moves k + 1 points and let a 1 A be a point moved by ρ. Define a 2 = ρ (a 1, a 3 = ρ (a 2,.... The list a 1, a 2, a 3,... consists of elements from A, a finite set, and so the list must repeat, i.e. there is a smallest n such that there exists 1 m < n with ρ (a n = a m. a m-1 a m a m+1 a n a 3 a n-1 a 2 a 1 Assume for a contradiction that m > 1. We can see from the diagram that a m is the image of two elements, i.e. a m = ρ (a m 1 and a m = ρ (a n. Thus ρ (a n = ρ (a m 1. Yet ρ is injective (1 1, so a n = a m 1. But a n has be defined to be ρ (a n 1 and so we have ρ (a n 1 = a n = a m 1, i.e. ρ (a n 1 = a m 1. But this contradicts the choice of n as the smallest positive value such that ρ (a n = a m for some m. Thus our last assumption is false and we must have m = 1, i.e. ρ (a n = a 1. Hence we have a cycle (a 1, a 2,..., a n which we label σ. Since this cycle moves a 1 it is non-trivial and so n 2. Note that if σ moves an element a then, by definition, it sends it to the same place as ρ sends it, in particular this means that σ only moves n of the k + 1 elements moved by ρ. But further the fact that if σ moves an element a then it sends it to the same place as ρ sends it, which can be written as (σ (a a (σ (a = ρ (a, has 1

Let ρ be a permutation that moves k + 1 points and let a 1 A be a point moved by ρ. Define a 2 = ρ (a 1, a 3 = ρ (a 2,.... The list a 1, a 2, a 3,.

2 the contrapositive σ (a ρ (a σ (a = a. (1 Consider the permutation σ 1 ρ. If σ 1 ρ moves an element a, i.e. σ 1 ρ (a a, then σ ( σ 1 ρ (a σ (a since σ is bijective and thus injective, σ ( σ 1 ρ ( (a σ (a σ σ 1 ρ (a σ (a 1 A ρ (a σ (a ρ (a σ (a σ (a = a by (1. Thus if a is moved by σ 1 ρ it is fixed by σ. We can combine the last two lines in the list, ρ (a σ (a and σ (a = a to get ρ (a a, i.e. ρ moves a. Hence if a is moved by σ 1 ρ it is moved by ρ but fixed by σ. These elements are k + 1 n in number. Since n 2 this is k 1. So by our inductive hypothesis σ 1 ρ can be expressed as a product of disjoint cycles, i.e. σ 1 ρ = σ 1 σ t. The elements moved by σ 1 σ t are fixed by σ and so σ 1 σ t and σ are disjoint, and thus σ 1,, σ t and σ are all disjoint. Finally ρ = σ σ 1 σ t as required. Example From above we have ( π = = (1, 5, 4 (2, 6 = (2, 6 (1, 5, We can go further and prove that the decomposition into disjoint cycles is unique. A proof of this is by induction and within it you need to be able to cancel permutations. Cancelation Law Assume that α, β and γ are permutations on a set A. If γ α = γ β then α = β. If α γ = β γ then α = β. 2

Hence if a is moved by σ 1 ρ it is moved by ρ but fixed by σ. These elements are k + 1 n in number. Since n 2 this is k 1.

3 Proof Assume γ α = γ β. Then α = 1 A α = ( γ 1 γ α = γ 1 (γ α composition is associative = γ 1 (γ β by assumption = ( γ 1 γ β = 1 A β = β. Leave the other case to students. Theorem A permutation on a finite set A can be expressed as a product of disjoint cycles uniquely apart from the order of the cycles. Proof Not given, but see the appendix. Definition The positive powers ρ n of a permutation are defined inductively by setting ρ 1 = ρ and ρ k+1 = ρ ρ k for all k N. The negative powers of a permutation are defined by ρ n = (ρ 1 n for all n N. Finally, we set ρ 0 = 1 A. Theorem Let ρ be a permutation of a finite set A. Then there is an integer m 1 such that ρ m = 1 A. Proof ρ, ρ 2, ρ 3,... is a list of bijections from the finite set A to itself. The number of such bijections is finite (since A is finite and so this list must repeat, i.e. there exist r < s such that ρ r = ρ s. Apply ρ r to both sides to get 1 A = ρ s r. Take m = s r. Definition The order or period of a permutation ρ of a finite set is the least positive integer n such that ρ n = 1 A. Examples In S 4 consider Then π 2 1 = π 1 = and so the order is 2. But consider π 2 = = 1 4,,

Definition The positive powers ρ n of a permutation are defined inductively by setting ρ 1 = ρ and ρ k+1 = ρ ρ k for all k N.

4 then π 2 2 = π 3 2 = π 2 π 2 2 =, = 1 4, and so the order is 3. But what about finding the order of something a little larger? In S 7 consider ( π = Then π 2 = , π 3 = ,... How long do we have to go on for? What if we had a permutation from S 100? Question For what permutations is it easy to calculate the order? The answer is cycles, but to prove this we need a result that uses the division algorithm seen earlier in the course. Theorem If the order of π is d then π e = 1 A if, and only if, d e. Proof ( Assume π e = 1 A. By the division Algorithm write e = qd + r for some integers q and 0 r d 1. Then 1 A = π e = π qd+r = ( π d q π r = (1 A q π r = π r. But d is the least positive integer with π d = 1 A and so r = 0. That is, e = qd and so d e. ( Assume d e. So e = dq for some q Z. But then π e = ( π d q = (1A q = 1 A. Theorem The order of a cycle is equal to its length. Proof Not given, but see the appendix. 4

Question For what permutations is it easy to calculate the order? The answer is cycles, but to prove this we need a result that uses the division algorithm seen earlier in the course.

5 Definition The lowest common multiple of integers m 1, m 2,..., m t, denoted by lcm (m 1, m 2,..., m t is the positive integer f that satisfies (1 m 1 f, m 2 f,..., m t f, (2 if m 1 k, m 2 k,..., m t k then f k. In words, (1 says that f is a common multiple of the integers, while (2 says that it is the least of all possible positive common multiples. Compare the definition to that of gcd. Theorem Suppose that π = π 1 π 2... π m is a decomposition into a product of disjoint permutations, then the order of π is the least common multiple of the orders of the permutations π 1, π 2,..., π m. Proof not given but see the appendix. Note In practice, given a permutation π we decompose it into a product of disjoint cycles. The order of each cycle is its length, so the order of π is the lowest common multiple of the lengths of the cycles. Since the decomposition into cycles is unique, apart from ordering, the lowest common multiple is well-defined. Examples In S 12 consider ( π = = (4, 10, 8, 9, 7 (2, 3, 5 (1, 6 (11, 12. The order equals lcm (5, 3, 2, 2 = 30. Example What is the largest order of all permutations in S 12? Solution Need to find positive integers a, b, c,... that sum to 12 but for which lcm (a, b, c,.. is as large as possible. Just search to find 12 = , when lcm (3, 4, 5 = 60. So, for example (1, 2, 3 (4, 5, 6, 7 (8, 9, 10, 11, = has order 60. 5

Theorem Suppose that π = π 1 π 2... π m is a decomposition into a product of disjoint permutations, then the order of π is the least common multiple of the orders of the permutations π 1, π 2,..., π m.

6 Example S 8. What is the order of (1, 2, 4, 6, 8 (2, 3, 6 (6, 7? CAREFUL, the cycles are not disjoint! We have to write this as a product of disjoint cycles. In fact it equals (1, 2, 3, 8 (4, 6, 7, now a composition of disjoint cycles. The order is lcm (4, 3 = 12. 6

Lemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S.

Lemma 5.2. Let S be a set. (1) Let f and g be two permutations of S. Then the composition of f and g is a permutation of S. Definition 51 Let S be a set bijection f : S S 5 Permutation groups A permutation of S is simply a Lemma 52 Let S be a set (1) Let f and g be two permutations of S Then the composition of f and g is a