Math 375 Week 6 6.1 Homomorphisms and Isomorphisms DEFINITION 1 Let G and G groups and let : G ! G b e a function. Then is a 1 2 1 2 group homomorphism if for every a; b 2 G wehave ab= ab: REMARK 1 Notice that the op eration on the left is o ccurring in G while the op er- 1 ation on the right is o ccurring in G . 2 REMARK 2 Notice the similarity to the de nition of a linear transformation from Math 204. I encourage you to lo ok this up in your 204 text. This means that you can multiply b efore or after you apply the mapping and you will still get the same answer. This is great, you can't make a mistake here b ecause the order of op erations mapping versus multiplication do es not matter. EXAMPLE 1 Consider the following maps a Is the mapping : GLn; R ! R by A = det A a homomor- phism? Yes. 1 b Let a b e a xed elementofG. Is : G ! G by g = ag a a homomorphism? [Homework] x c Is the mapping f : R ! R by f x= e a group homomorphism? Be careful: What are the group op erations in each case? d Let a b e a xed elementofG. Is : G ! G by g = ag a homomorphism? No. e Here's a silly example: Let G and G groups and let : G ! G 1 2 1 2 by g = e for all g 2 G.Obviously, ab= e = e e = bb, 2 2 2 2 so this is a group homomorphism. f Recall that S is the set of all p ermutations of the set By lab elling 3 the vertices of an equilateral triangle 1, 2, and 3 as usual, we can interpret the elements of D as maps from S to S . Match up the 3 elements of D with their corresp onding elements in S . 3 3 1 2 Math 375 EXAMPLE 2 Extra Credit: Lab el the vertices of a tetrahedron with 1, 2, 3, 4. Let G b e the set of rotations and re ections of the tetrahedron. Write out each such rotation as an elementofS .Doyou get all elements of S 4 4 this way? See page 101 and 102 in your text. It turns out that this pairing is a group homomorphism, indeed, an isomorphism. THEOREM 2 Basic Prop erties of Homomorphisms If : G ! G is a group 1 2 homomorphism, then a e = e ; 1 2 1 1 b a = [a] ; n n c a = [a] for all n 2 Z; d if jaj = n, then jaj n, i.e., jaj jaj. PROOF A Note that e =e e = fe e so that by cancellation e = 1 1 1 1 1 2 e . [Rememb er, this is all taking place in G .] 1 2 PROOF B We prove that something is an inverse by showing that it acts likean inverse. So 1 1 e = e = aa = aa : 2 1 1 1 1 So f a acts as the inverse to f a, i.e., a = [a] . PROOF C Homework. n PROOF D Because jaj = n, then a = e .So 1 n n e = e = a =[a] : 2 1 By the Corollary on page 73, jaj n. EXAMPLE 3 Supp ose that : Z ! D is a homomorphism. Can 1 = r ? 3 4 90 SOLUTION No. Because the last part of the theorem we need j1j j1j or but jr j =46j3= j1j. 90 EXAMPLE 4 Let's continue with : Z ! D . What can you say ab out this homo- 3 4 morphism? SOLUTION Since the orders of elements in D are either 1, 2, or 4, the only such 4 order which divides j1j = 3 is 1. So, 1 = r . But then part b of the 0 1 1 thero em ab ove, 2 = 1 = [1] = r = r . Of course, by part 0 0 a of the theorem, 0 = r .Soevery elementinZ must b e mapp ed 0 3 to r . This is the silly example discussed ab ove. 0 Let's prove something a bit more interesting ab out homomorphisms of nite cyclic groups. Supp ose that G =<a>is cyclic and : G ! H 6.1 Homomorphisms and Isomorphisms 3 is a group homomorphism. Notice that is completely determined by k where maps a. Because any element g 2 G is of the form g = a ,so k k g = a = [a] . Once we know what a is, we know what is. What are the choices for a? EXAMPLE 5 Her's the sort of thing I mean. Let's assume : Z ! Z is a homomor- 8 4 phism. Supp ose that 1 = 3 . Then the rest of is now completely 8 4 determined b ecause Z =< 1 >.We b ecause is a homomorphism 8 8 1 ! 3 8 4 2 = 1 +1 ! 3 +3 =2 8 8 8 4 4 4 3 = 1 +2 ! 3 +2 =1 8 8 8 4 4 4 4 = 1 +3 ! 3 +1 =0 8 8 8 4 4 4 5 = 1 +4 ! 3 +0 =3 8 8 8 4 3 4 6 = 1 +5 ! 3 +3 =2 8 8 8 4 4 4 7 = 1 +6 ! 3 +2 =1 8 8 8 4 4 4 0 = 1 +7 ! 3 +1 =0 8 8 8 4 4 4 Or one could use j = j 1 ! j 3 to get the same answers. 8 8 4 Ok, let's generalize nite case rst. LEMMA 3 Let G =<a>b e a cyclic group of order n. Let : G ! H be a i i function such that a =a for all i.Ifjaj n, then is a group homomorphism. PROOF Note from parts c and d of the previous theorem, if is a homomor- phism, then these two prop erties must b e true. This says that these two prop erties suce to make a homomorphism when G is cyclic. Let jaj = m. Then we are given that m j n,son = md for some d 2 Z. Let x; y 2 G.Wemust show that xy = xy . But G j k j k is cyclic so x = a and y = a with 0 k;j<n. Then xy = a a = j +k mo d n a .So j +k mo d n j +k mo d nmodm xy = a =[a] : The mo d m is necessary since jaj = m. On the other hand, j k j mo d m k mo d m j +k mod m xy = a a =[a] [a] =[a] : So it all b oils down to whether j + k mo d nmodm =j + k mod m. By the division algorithm, wemay write j + k = qn + s 0 s< n: 4 Math 375 So j + k mo d nmodm = s mo d m Since n = dm,wehave j + k = qn + s = q dm+ s =qdm + s: But then j + k modm = s mo d m, to o. So j +k mo d nmodm s mo d m xy = [a] =[a] j +k mod m =[a] j mo d m k mo d m =[a] [a] = xy : When G is an in nite cyclic group the pro of is even easier. LEMMA 4 Let G =<a> b e an in nite cyclic group. Let : G ! H b e a function i i such that a =a for all i. Then is a group homomorphism. Again, let x; y 2 G.Wemust show that xy = xy . But G is j k cyclic so x = a and y = a There are two cases. If jaj = 1, then by assumption j +k j +k j k xy = a = [a] =[a] [a] = xy : If jaj = n, then by assumption j +k j +k mod n xy = a = [a] j mo d n k mo d n =[a] [a] = xy : Another crucial fact is that comp osites of homomorphisms are ho- momorphisms. LEMMA 5 Let : G ! H , and : H ! K b e group homomorphisms. Then so is the comp osite, : G ! K . PROOF Let a; b 2 G. Then since b oth maps are homomorphisms, ab= ab = ab = a f b= [ a][ b]: DEFINITION 6 If in addition a homomorphism : G ! G is b oth injective and 1 2 surjective then is called a group isomorphism. The two groups are said to b e isomorphic and this is denoted by G G . = 1 2 REMARK Note that to provetwo groups are isomorphic, wemust 1 nd a map- ping : G ! G ; 2 show that is injective; 3 show that is 1 2 surjective; and 4 show that is a homomorphism. 6.1 Homomorphisms and Isomorphisms 5 EXAMPLE a For homework, if G is a group and a is a xed elelmentofG, then 1 the mapping : G ! G by g = ag a is an injective, surjective homomorphism. Thus is an isomorphism. b Wesaw that if G were a group and a was a xed elelmentofG, then the mapping : G ! G by g = ag was an injective and surjective. Let's check to see if it is an isomorhism. gh = ag h, while gh= ag ah. The two are not equal, and thus is not an isomorphism. c The simplest example is the identity mapping. Let G b e a group and let i : G ! G by i g = g .We know that i is injective and G G G surjective and clearly i ab= ab = i ai b: G G G So i is an isomorphism and G G.