Ma/CS 6a Class 19: Group Isomorphisms

Transcription

1 Ma/CS 6a Class 19: Group Isomorphisms By Adam Sheffer A Group A group consists of a set G and a binary operation, satisfying the following. Closure. For every x, y G x y G. Associativity. For every x, y, z G x y z = x y z. Identity. There exists e G, such that for every x G e x = x e = x. Inverse. For every x G there exists x 1 G such that x x 1 = x 1 x = e. 1

2 Reminder: What We Already Know About Groups Given a group with a set G: For a, b G, the equation ax = b has a unique solution. The identity is unique. For each a G, there is a unique inverse a 1. The multiplication table of G is a Latin square. Reminder: Orders The order of a group is the number of elements in its set G. The order of an element a G is the least positive integer k that satisfies a k = Rotation 90 (under multiplication mod 3) (under integer addition) 4 2 2

3 Symmetries of a Tiling Given a repetitive tiling of the plane, its symmetries are the transformations of the plane that Map the tiling to itself (ignoring colors). Preserve distances. These are combinations of translations, rotations, and reflections. Example: Square Tiling What are the symmetries of the square tiling? Translations in various directions. Rotations around a vertex by 0, 90, 180, 270. Rotations around the center of a square by 0, 90, 180, 270. Reflections across vertical, horizontal and diagonal lines. Rotations around the center of an edge by 180. Translation + reflection. 3

4 Wallpaper Groups Given a tiling, its set of symmetries is a group called a wallpaper group (not accurate. We ignore an extra condition). Closure. Composing two symmetries results in a transformation that preserves distances and takes the tiling to itself. Associativity. Holds. Identity. The no operation element. Inverse. Since symmetries are bijections from the plane to itself, inverses are well defined. Wallpaper Groups There are exactly 17 different wallpaper groups. Every repetitive tilings of the plane belongs to one of these 17 classes. Two tilings of the same class have the same symmetries. 4

5 Reminder: A Group of 2 2 Matrices 2 2 matrices of the form α β where α 1,2 and β 0,1,2. The operation is matrix multiplication mod 3. The group is of order 6: The Multiplication Table I = 1 0 X = 2 0 R = 1 1 Y = 2 1 S = 1 2 Z = 2 2 I R S X Y Z I I R S X Y Z R R S I Y Z X S S I R Z X Y X X Z Y I S R Y Y X Z R I S Z Z Y X S R I 5

6 Symmetries of a Triangle The six symmetries of the equilateral triangle form a group (under composition). Another Multiplication Table i r s x y z i r s x y z i i r s x y z r r s i y z x s s i r z x y x x z y i s r y y x z r i s z z y x s r i 6

7 i r s x y z i i r s x y z r r s i y z x s s i r z x y x x z y i s r y y x z r i s z z y x s r i I R S X Y Z I I R S X Y Z R R S I Y Z X S S I R Z X Y X X Z Y I S R Y Y X Z R I S Z Z Y X S R I Isomorphisms G 1, G 2 two groups of the same order. A bijection β: G 1 G 2 is an isomorphism if for every a, b G 1, we have β ab = β a β b. ( = after reordering, we have the same multiplication tables) When such an isomorphism exists, G 1 and G 2 are said to be isomorphic, and we write G 1 G 2. 7

8 Isomorphism Over the Reals Problem. Prove that the following groups are isomorphic: The set of real numbers R under addition. The set of positive real numbers R + under multiplication. Proof. Use the functions e x : R R + and log x : R + R as bijections between the two sets. For x, y R we have e x+y = e x e y. For x, y R +, we have log xy = log x + log y. Isomorphism Between Z 4 and Z 5 +? Problem. Are the following two groups isomorphic? The set Z 4 = {0,1,2,3} under addition mod 4. The set Z 5 + = 1,2,3,4 under multiplication mod 5. Solution. Yes. Use the following bijection of Z 4 Z

9 An Isomorphism? Are the following two groups isomorphic? Nonzero real numbers under multiplication. Rational numbers under addition. What would such an isomorphism satisfy? The identity 1 corresponds to the identity 0. In both cases most elements have an infinite order. While 1 2 is the identity, no nonzero rational satisfies a + a = 0. Not isomorphic! Cyclic Groups A group G is cyclic if there exists an element x G such that every member of G is a power of x. We say that x generates G. What is the order of x? G. An infinite group G is cyclic if there exists an element x G such that G =, x 2, x 1, 1, x, x 2, x 3, 9

10 Cyclic Groups? Are the following groups cyclic? Integers under addition. Yes! It is generated by the integer 1. The symmetries of the triangle. No. There are no generators. Cyclic Groups? Are the following groups cyclic? The positive reals R + under multiplication. No. There is no generator. The aforementioned group of elements of the α β form. No. Because it is isomorphic to the triangle symmetry group. 10

11 Finite Cyclic Groups Any cyclic group of a finite order m with generator g can be written as 1, g, g 2,, g m 1. For integers q and 0 r < m, we have g qm+r = g r. Where did we already encounter such a group? Integers mod m under addition. The generator is 1 and the group is 0,1,2,, m 1. Isomorphic Cyclic Groups Claim. All of the cyclic groups of a finite order m are isomorphic. We refer to this group as C m. Proof. Consider two such cyclic groups G 1 = 1, g, g 2, g 3,, g m 1, G 2 = 1, h, h 2, h 3,, h m 1. Consider the bijection β: G 1 G 2 satisfying β g i = h i. This is an isomorphism since β g i g j = β g i+j = h i+j = h i h j = β g i β g j. 11

12 Direct Product G 1, G 2 - two groups with identities 1 1, 1 2. The direct product G 1 G 2 consists of the ordered pairs a, b where a G 1 and b G 2. The direct product is a group: The group operation is a, b c, d = ac, bd. The identity element is 1 1, 1 2. The inverse a, b 1 is a 1, b 1. The order of G 1 G 2 is G 1 G 2. Direct Product Example Problem. Is C 6 isomorphic to C 2 C 3? Solution. Write C 2 = 1, g and C 3 = 1, h, h 2. Then C 2 C 3 consists of 1,1, 1, h, 1, h 2, g, 1, g, h, (g, h 2 ). C 2 C 3 is isomorphic to C 6 iff it is cyclic. It is cyclic, since it is generated by g, h. g, h 1 = g, h, g, h 2 = 1, h 2, g, h 3 = g, 1, g, h 4 = 1, h, g, h 5 = g, h 2, g, h 6 = 1,1, 12

13 Another Direct Product Example Problem. Is C 8 isomorphic to C 2 C 4? Solution. Write C 2 = 1, g and C 4 = 1, h, h 2, h 3. Then C 2 C 4 consists of { 1,1, 1, h, 1, h 2, 1, h 3, g, 1, g, h, g, h 2, g, h 3 }. C 2 C 4 is isomorphic to C 8 iff it is cyclic. It is not. The orders of the elements are 1,4,2,4,2,4,2,4, respectively. Cyclic Inner Products of Cyclic Groups Claim. If m, n are relatively prime positive integers, then C m C n C mn. Proof. Write C m = 1, g, g 2,, g m 1 and C n = 1, h, h 2,, h n 1. It suffices to prove that g, h generates C m C n. g, h k = 1,1 if and only if m k and n k. GCD m, n = 1 implies LCM m, n = mn (see example assignment). Thus, the order of g, h is mn. 13

14 Cyclic Inner Products of Cyclic Groups Claim. If m, n are relatively prime positive integers, then C m C n C mn. Proof. Write C m = 1, g, g 2,, g m 1 and C n = 1, h, h 2,, h n 1. We proved that g, h is of order mn. It remains to show that for every 0 i < j < mn, we have g, h i g, h j. If g, h i = g, h j, multiplying both sides by g, h i implies g, h j i = 1,1. Contradiction to the order of g, h! So g, h generates mn distinct elements. The End: Alhambra Alhmbra is a palace and fortress complex located in Granada, Spain. The Islamic art on the walls is claimed to contain all 17 wallpaper groups. Mathematicians like to visit the palace and look for as many types as they can find. 14