301 Abstract Algebra 2:00 - 2:50 MWF Hentzel 432 Carver hentzel@iastate.edu Web Page http://www.math.iastate.edu/hentzel/class.301 Text: A First Course in Abstract Algebra, sixth Edition by John B. Fraleigh Course: Chapter 1,2,3,4,5 The material we are covering today and Friday is on Page 197 to 208 Section 2.3 beginning with "Group Action on a Set." On Friday we will do the theory which is essentially what is given in the text. Today we give the application. This is not covered very well in the text as far as I can tell, but it is what he is trying to present in the five examples on pages 206-207. Main Idea: Counting the various distinct objects. Key Words: Fixed. Objectives: Counting the number of distinctly colored cubes, necklaces, and place mats. If one has a bracelet with six beads on it. Then the following two bracelets are the same X-------0 0-------0 / \ / \ / \ / \ 0 0 0 X \ / \ / \ / \ / 0-------0 0-------0 But the following two bracelets are different. X-------X X-------0 / \ / \ / \ / \ 0 0 0 X \ / \ / \ / \ / 0-------0 0-------0 The reason is that one can twist the first into the second in the top case, but no amount of twisting will deform the bracelet so that positions which are together will move apart. In a situation like this, we are given a group which are the actions allowed. If two objects can be adjusted by the group till they are identical, then we call those objects identical. The problem is to count how many different objects there are. This is a numerical lecture. The proofs that what you are doing really works will come on Friday. How many distinct bracelets with 6 beads can be constructed using 2 colors of beads. 1-------2 / \ / \ 6 3 \ / \ / 5-------4 The group I (123456) (135)(246) (14)(25)(36) (153)(264) (165432) corner flips 3 of them of type ( )( )( , )( , ) edge flips 3 of them of type ( , )( , )( , ) left fixed The group has I 1 64 = 64 2 ( , , , , , ) 2 2 4 2 ( , , )( , , ) 2 4 8 3 ( )( )( , )( , ) 3 16 48 4 ( , )( , )( , ) 4 8 32 ----------------------- -------- 12 elements total. 156 Number of necklaces is 156/12 = 13. Once the bracelet is on the wrist, how many are there? Group is: I (123456) (135)(246) (14)(25)(36) (153)(264) (165432) Fixed: 64 2 4 8 4 2 84/6 = 14 This means that one bracelet has two distinct ways to wear it, depending on which way it is slipped over the arm. Which is it? ##################### #...................# #.........|.........# #...red...|...blue..# #.........|.........# #_________|_________# #.........|.........# #..green..|...blue..# #.........|.........# #.........|.........# ##################### How many place mats are there using 3 colors? Symmetries of the Square I ( )( )( )( ) 81 2 ( , , , ) 6 2 ( , )( )( ) 54 3 ( , )( , ) 27 ------------- ----- 8 168 168/8 = 21 What are they. Same color 3-1 2-2 1-1-2 AA AA AA AB A B A C AA AB BB BA C C C B 3 6 3 3 3 3 What is the group of symmetries of a Cube? How many elements are there and what are their cycle structure. Problems: (1) How many ways can you paint the faces of a cube with 2 colors? (2) How many ways can you paint the faces of a cube with 3 colors? (3) How many 8 bead necklaces are there with 2 colors? (4) How many 8 bead necklaces are there with 3 colors?