6 Permutations Very little of this section comes from PJE.

Size: px

Start display at page:

Download "6 Permutations Very little of this section comes from PJE."

Alice Jordan
5 years ago
Views:

1 6 Permutations Very little of this section comes from PJE Definition A permutation (p147 of a set A is a bijection ρ : A A Notation If A = {a b c } and ρ is a permutation on A we can express the action of ρ on A using a two row notation due to Cauchy: ( a b c ρ = ρ (a ρ (b ρ (c Note The identity map 1 A which satisfies 1 A (α = α for all α A is therefore given by ( a b c 1 A = a b c Definition The collection of all permutations on a set A denoted by S A is called the symmetric group on A When A = {1 2 3 n} S A is usually denoted by S n and is called the symmetric group on n letters Let 1 n denote the identity map in S n Example S 3 consists of Theorem If A is a finite set then f : A A is a bijection if and only if it is an injection Proof p138 and will not be covered in my part of the course Corollary If A = n 1 then the number of permutations ρ : A A is n! Proof Since A is a finite set to count bijections it suffices to count injections Hence S A = Inj (A A = n! Recall if ρ and π are functions A A then the composite function is defined by ρ π (a = ρ (π (a for all a A Further if ρ and π are bijections then ρ π is a bijection Hence the composition of permutations is a permutation Example Let ρ π S 5 be given by ( ρ = and

2 To write ρ π in the same way we have to see first what π does to a given element of A and then secondly what ρ does to this image In this example π (1 = 2 ρ (2 = 2 so ρ π (1 = 2 π (2 = 3 ρ (3 = 1 so ρ π (2 = 1 π (3 = 4 ρ (4 = 3 so ρ π (3 = 3 π (4 = 5 ρ (5 = 5 so ρ π (4 = 5 π (5 = 1 ρ (1 = 4 so ρ π (5 = 4 Thus ρ = ( ( ( Note we first looked at π then at ρ so read ρ π from the right Note also that π ρ = so that for instance ρ π (1 = 2 but π ρ (1 = 5 Thus ρ π π ρ and hence composition of permutations is not commutative Recall a bijection always has an inverse The inverse of a permutation written in the two row manner can easily be found by exchanging upper and lower rows and then reordering the columns so the entries on the upper row appear in the same order as in the original permutation Examples In S 5 the inverse of is ( = You should check that your answer satisfies the definition of inverse namely that f f 1 = f 1 f = 1 ( ( ( ( = =

3 Definition If ρ is a permutation on A then ρ fixes a A if ρ (a = a and ρ moves a if ρ (a a Definition Let a 1 a 2 a r be elements in A If ρ is a permutation that fixes all the other elements of A and if ie ρ (a 1 = a 2 ρ (a 2 = a 3 ρ (a 3 = a 4 ρ (a r 1 = a r ρ (a r = a 1 a 1 a 2 a 3 a r a 1 then ρ is called a cycle of length r sometimes called an r-cycle A 2-cycle is called a transposition The r-cycle above will be denoted by (a 1 a 2 a 3 a r Note that any a i can be taken as the starting point so (a 1 a 2 a 3 a r = (a 2 a 3 a r a 1 = = (a r a 1 a r 2 a r 1 We can take r = 1 in the definition to get a 1-cycle (a 1 But such a cycle fixes all elements of A is thus the identity Hence all 1-cycles equal the identity ie (a = 1 A for all a A Examples (i Two permutations seen before were cycles Namely ρ π S 5 ( ρ = = ( and So = ( (ii In S 3 all permutation happen to be cycles namely 1 3 (2 3 (1 2 (1 3 (1 3 2 and (1 2 3 The inverse of a cycle is obtained simply by writing it in reverse order ( ρ 1 = ( = (3 4 1 = as seen before And we can compose cycles written in this notation remembering to read from the right So ( ρ (1 4 3 ( =

4 as seen before Again we did this by noting that π moved 1 to 2 which ρ then fixed Next π moved 2 to 3 which ρ moved onto 1 Continue Example In S 3 we can represent all possible products in a table 1 3 (2 3 (1 2 (1 3 (1 3 2 ( (2 3 (1 2 (1 3 (1 3 2 (1 2 3 (2 3 ( (1 3 2 (1 2 3 (1 2 (1 3 (1 2 (1 2 ( (1 3 2 (1 3 (2 3 (1 3 (1 3 (1 3 2 ( (2 3 (1 2 (1 3 2 (1 3 2 (1 3 (2 3 (1 2 ( (1 2 3 (1 2 3 (1 2 (1 3 ( (1 3 2 Note that because composition is not commutative this table is not symmetric about it s leading diagonal But is does have other nice properties: it is closed and every row and every column has all the elements of S 3 in some order Question If we can multiply permutations can we factor them? Algorithm for factorization is best illustrated by an example Example In S 6 consider Take an element in { } for example 1 See what π does to 1 It sends 1 to 5 Then π sends 5 to 4 Next π sends 4 back to 1 Thus we have a cycle (1 5 4 Next look at any element not in this cycle 2 say Then we see that and so we get another cycle (2 6 Repeat by taking any element not in these two cycles We have only one such element 5 and this is fixed by π and so we get a 1-cycle (5 which we know is the identity When there is at least one non-identity cycle we can omit the identity (5 Question In which order do we multiply these cycles ie do we have (1 5 4 (2 6 or (2 6 (1 5 4? Definition Two permutations ρ and π of a set A are disjoint if every element moved by ρ is fixed by π and every element moved by π is fixed by ρ (In other words the sets of objects moved by each permutation are disjoint sets Example In S 5 the cycles (1 5 4 and (2 6 are disjoint The cycles (1 4 3 and ( are not disjoint 4

5 a m a m+1 Theorem Disjoint permutations on a set commute Proof not given in course (see Additional Notes on the web site Theorem A permutation on a finite set A is either a cycle or can be expressed as a product 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 define P A to be the set of points moved by ρ ie P = {a A : ρ (a a} Then P = k + 1 Let a 1 P and 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 ie there is a smallest n such that there exists 1 m < n with ρ (a n = a m a m-1 a 3 a n 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 ie 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 ie 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 ie 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 5

6 Let S be the set of elements moved by σ so S = {a A : σ (a a} Then S = n If a S then a = a i for some 1 i n Since σ is a cycle σ (a i = a i+1 with the convention that a n+1 = a 1 Yet by the definition of σ we have a i+1 = ρ (a i Hence σ (a i = a i+1 = ρ (a i ie σ (a i = ρ (a i or σ (a = ρ (a for all a S In particular for a S ρ (a = σ (a a and so a P and thus S P Consider σ 1 ρ If a S P Then σ (a = ρ (a and so σ 1 ρ (a = a ie a is fixed by σ 1 ρ Let a P \ S then ρ (a a while σ (a = a Thus ρ (a σ (a and so σ 1 ρ (a a ie a is moved by σ 1 ρ Finally if a / P (in which case a / S then ρ (a = a and σ (a = a Thus σ (a = ρ (a and so σ 1 ρ (a = a ie a is fixed by σ 1 ρ Combining these three observations we conclude that a is moved by σ 1 ρ if and only if a P \ S Since S P we have P \ S = k+1 n ie σ 1 ρ moves k+1 n k 1 elements So by our inductive hypothesis σ 1 ρ can be expressed as a product of disjoint cycles ie σ 1 ρ = σ 1 σ t Thus ρ = σ σ 1 σ t as required Example From above we have ( = (1 5 4 (2 6 = (2 6 ( We can go further 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 in course (see Additional Notes on the web site 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 where ρ 1 is the inverse of ρ Finally we set ρ 0 = 1 A 6

7 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 ie 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 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 4 π 2 = = 1 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 ( Then π 2 = π 3 = How long do we have to go on for? What if we had a permutation from S 100? Theorem If the order of π is d then π e = 1 A if and only if d e 7

8 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 Question For what permutations is it easy to calculate the order? Theorem The order of a cycle is equal to its length Proof not given in course (see Additional Notes on the web site 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 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 in course (see Additional Notes on the web site 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 lcm of the lengths of the cycles Since the decomposition into cycles is unique apart from ordering the lcm is well-defined Examples In S 12 consider ( = ( (2 3 5 (1 6 (

9 The order equals lcm ( = 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 ( ( = has order 60 Example S 8 What is the order of ( (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 ( (4 6 7 now a composition of disjoint cycles The order is lcm (4 3 = 12 Appendix We have seen that we can build all permutations out of cycles But there are other possible building blocks Definition A transposition is a cycle of length 2 Example (2 3 S 5 is a transposition Example In S 5 we have (2 3 4 = (3 4 (2 4 ie we have written a permutation as a product of transpositions This decomposition is not unique for example (2 3 4 = (3 4 2 = (4 2 (3 2 and the sets of transpositions {(4 2 (3 2} {(3 4 (2 4} are different But we do have a general result Theorem If A 2 then every cycle on A is a product of transpositions Proof If π is a 1-cycle then π is the identity which can be written as (a 1 a 2 (a 2 a 1 say If π is an r-cycle with r 2 then (a 1 a 2 a r = (a 1 a r (a 1 a r 1 (a 1 a 3 (a 1 a 2 9

10 Students to check Example In S 4 ( = (1 4 (1 3 (1 2 = (1 4 (2 3 (1 3 so this form of decomposition is not unique But from the example in S 4 (1 2 3 = (1 3 (1 2 = (1 3 (4 2 (1 2 (1 4 = (1 3 (4 2 (1 2 (1 4 (2 3 (2 3 it might not seem unreasonable that the parity of number of factor (ie odd or even is the same for all factorizations We do not pursue this further Corollary A permutation can be expressed as a product of transpositions Proof Immediate on combining the above results Example In S 6 we have ( = (1 5 4 ( = (1 4 (1 5 (2 6 Example In S 7 we have ( = ( ( = (6 5 (3 5 (1 5 (7 4 (2 4 Of course this decomposition is neither unique or disjoint 10

Definition 6.1 (p.277) A positive integer n is prime when n > 1 and the only positive divisors are 1 and n. Alternatively

6 Prime Numbers Part VI of PJE 6.1 Fundamental Results Definition 6.1 (p.277) A positive integer n is prime when n > 1 and the only positive divisors are 1 and n. Alternatively D (p) = { p 1 1 p}. Otherwise