Math 375 Week 5 5.1 Symmetric Groups The Group S X Now let X be any non-empty set. Let S = fj : X ! X and f is b oth surjective and injective g: X Notice that if ; 2 S , then : X ! X and is surjective and injec- X tiveby previous arguments. That is, the comp osition 2 S .Thus X comp osition is a binary op eration on S . X THEOREM 1 S is a group under comp osition. S is called the Symmteric Group X X on X . PROOF i We just checked closure. ii Asso ciativity: mentioned last time see text Chapter 0. iii The identityinS is i b ecause X x= X X i x = x, that is, i = . Similarly i = . iv Evidently X X X 1 the inverse of is whichwe know exists b ecause is injective and 1 1 surjective and x= x = i x and x= x = i x. X X This last example shows just how general the group concept is. Sym- metric groups are extraordinarily imp ortant. Arthur Cayley as in Cay- ley table showed that every group is the subgroup of some symmetric group. So if you understand symmetric groups completely, then you un- derstand all groups! We can examine S for any set X .For example X if X = R, then examples of of elements in S are i , f : R ! R by R R a ! a +1, g : R ! R by a ! a=2, and so on. It is clear that S is R in nite. The individual elements of S are often called p ermutations of X the elements of X . This will b e the next big topic wecover. 1 2 Math 375 The Symmetric Group on n Elements: S n To make matters simpler, we will study symmetric groups of nite sets. For example, if X is a set of n elements, then wemayaswell lab el the elements of X as f1; 2;:::;ng.We usually denote the symmetric group on n elements by S . n Nowany elementorp ermutation in S is an injective and sur- n jective function from the set of the rst n integers to itself. It merely shues these elements around. Consequently it can b e represented as atworow matrix in which the rst row represents the input and the second represents the corresp onding output. EXAMPLE 1 List all the elements of S . 3 1 2 3 1 2 3 1 2 3 2 = ; = ; = 1 2 3 2 3 1 3 1 2 1 2 3 1 2 3 1 2 3 2 = ; = ; = 1 3 2 3 2 1 2 1 3 It should b e clear what the identity element is. Howdowe get the inverse of any element? Merely exchange rows, and re-order. How do we take the pro duct or comp osition of two such elements in S ? n In comp osing functions always remember to work from the right to the left whichisbackwards from what we read. This is a hold over from comp osition of functions, which is what we implicitly are doing here. 1 2 3 1 2 3 1 2 3 = 3 2 1 2 1 3 2 3 1 Notice that order matters: 1 2 3 1 2 3 1 2 3 = 2 1 3 3 2 1 3 1 2 Clearly S is ab elian, since it consits of only the identity element. 1 We'll see that S is ab elian in fact it's cyclic since it has only two 2 elements. However, wehave seen that S is not ab elian and in general: 3 THEOREM 2 If n 3 then S is non-ab elian. n 5.1 Symmetric Groups 3 PROOF Let and b e the following two elements in S : n 1 2 3 4 ::: n 1 2 3 4 ::: n = = : 3 2 1 4 ::: n 2 1 3 4 ::: n Then 1 2 3 4 ::: n 1 2 3 4 ::: n = = : 3 1 2 4 ::: n 2 3 1 4 ::: n EXAMPLE 2 Find the following inverse: 1 1 2 3 1 2 3 = 2 3 1 3 1 2 THEOREM 3 jS j = n!: n PROOF jS j is just the numberofways the integers 1 through n can arranged. n In other words in how many di erentways can we ll in the blanks: 1 2 ::: n ::: Well wehave n choices for the rst entry, and then n 1choices for the next entry, and so on yielding a total of n n 1 1= n! total choices. For example jS j =3!=6aswehave already seen. jS j = 2!=2. 3 2 If we write out the Cayley table for S we see that it is ab elian. Note 2 that the square of every element is the identity. 1 2 1 2 1 2 2 1 1 2 1 2 1 2 1 2 1 2 2 1 1 2 1 2 1 2 2 1 2 1 1 2 EXAMPLE 3 Compare the symmetries of an equilateral triangle to S . Write each 3 elementinD as a p ermutation. 3 EXAMPLE 4 Compare D to S . Are they the same group? 4 4 4 Math 375 Cycle Notation It will provemuch more convenient to use a di ernttyp e of notation to denote the individual p ermutations of S . This cycle notation is useful n b ecause it reveals the structure of individual elements in the group, much as p owers of a generator indicate the nature of a cyclic group. Here's howitworks. Consider the following elements of S : 5 1 2 3 4 5 1 2 3 4 5 = = 2 3 1 5 4 4 3 5 1 2 Notice that breaks quite naturally into two parts: 1 ! 2 ! 3 ! 1 and 4 ! 5 ! 4 Each of these pieces is called a cycle. The rst is a three-cycle b ecause it contains three di erent elements, the second is a two-cycle. Eachof these cycles can b e represented more compactly without the arrows: 1; 2; 3 and 4; 5 Let's lo ok at . This time the cycles are: 1 ! 4 ! 1 and 2 ! 3 ! 5 ! 2 or 1; 4 and 2; 3; 5 When using cycles we adopt the convention that if an elementis missing from the cycle, then it is mapp ed to itself. EXAMPLE 5 What p ermutation do es the cycle 2; 4; 3 representin S ? 6 1 2 3 4 5 6 2; 4; 3 = 1 4 2 3 5 6 EXAMPLE 6 Let =1; 3; 62; 4; 5 and =1; 2; 34; 5 in S . Then = 6 4; 2; 6; 153 = 4; 2; 6; 1. Do some more. EXAMPLE 7 What is the inverse of a single cycle: =4; 2; 3; 5? This is easy to sp ot in matrix form. If 1 2 3 4 5 =4; 2; 3; 5 = ; 1 3 5 2 4 then 1 2 3 4 5 1 = =5; 3; 2; 4: 1 4 2 5 3 More generally: 5.1 Symmetric Groups 5 THEOREM 4 The inverse of a cycle is the cycle in inverse reverse order. Further the inverse of a pro duct of cycles is the pro duct of the inverse cycles in reverse order. REMARK The second part of the theorem is the usual statement ab out inverses of pro ducts b eing the pro duct of the inverses in reverse order. DEFINITION 5 Two cycles are called disjoint if they contain no term in common. For example 1; 2; 6 and 3; 4 are disjoint but 1; 2; 4 and 3; 4; 5 are not. One of the basic facts ab out p ermutations is that THEOREM 6 Any p ermutation can b e written as a pro duct of disjoint cycles. The details of the pro of are presented in the text, but note that this is howwe rst got started on lo oking at cycles. Given an individual example of a p ermutation it is easy to see how to split it into disjoint cycles. Consider 1 2 3 4 5 6 7 8 9 : 5 9 3 8 6 1 4 7 2 We simply start the rst cycle with 1, continue backuntil we get 1. Then start the second cycle with the smallest remaining unused num- ber until we get back to it and so on. In this case the cycles are: 1; 5; 62; 934; 8; 7: THEOREM 7 Disjoint cycles commute. PROOF Let the disjoint cycles b e =a ;a ;:::;a and =b ;b ;:::; , 1 2 m 1 2 n where and havenoentries in common. Let the full set S b e given by S = fa ;a ;:::;a ;b ;b ;:::; ;c ;:::;c g: 1 2 m 1 2 n 1 k Wewant to prove that = .To do this, wemust show that a = x for all x 2 S . There are three cases: x is either an a, b,or c. Supp ose that x = a . Then i x= a = a = a = a : i i i+1 mo d m i+1 mo d m while x= a = a = a = a : i i i i+1 mo d m Hence the p ermutations agree on the a's. A similar argumentworks for the b's. The c's are the easiest of all. If x = c , then i x= c = c = c = c i i i i while x = c = c = c = c : i i i i Again the p ermutations are the same. 6 Math 375 5.2 More ab out S n The Order of a k -cycle EXAMPLE 8 Consider the following elements of S : =2; 3 and =1; 2; 4, and 5 =1; 3; 4; 2 Find the orders of each of these elements.