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 is on Page 120-125 Section 2.3 beginning with "Cosets and the Theorem of Lagrange." Main Idea: Subgroups chop up the group into identical sized pieces Key Words: Coset, Goal: Show that o(S) divides o(G) when S is a subgroup of G. Theorem (Lagrange page 124). If H is a subgroup of a finite group G, then |H| divides |G|. That is, the number of elements in H divides the number of elements of G. Proof: Suppose that the subgroup h = {h1,h2,h3, ..., hn}. We put the elements of H in the bottom of the stacking tray. Now we pick an element "a" of G which is not in H. We fill the next level by multiplying all the elements of H by a. Now we pick an element "b" of G which is not listed already and fill the next row my multiplying H by "b". And so on. As long as there are elements of G which have not yet appeared, we pick one and fill a new level by multiplying that element by the n elements of H. ------------------------ | | ------------------------ | | ------------------------ | | ------------------------ | | ------------------------ | h b, h b, ... h b | | 1 2 n | ------------------------ | h a, h a, ... h a | | 1 2 n | ------------------------ | h , h , ... h | | 1 2 n | ------------------------ The surprising occurrence is: (a) Each new row is n distinct elements. And (b) no element appears in two distinct rows. Since every element of G appears in one of the rows, we know that: | G | = (#of elements in H)*(number of rows) This means that the number of elements in H divides the number of elements in G. Or |H| divides |G|. The (number of rows) is called the index of H in G. Why (a)? Why does each new row have h distinct elements? Proof: Suppose that hi x = hj x where x was the element chosen to build the new row. Multiply by x' cancels off the x showing that hi = hj. This means that if all the hi's are distinct, then so are all of the hi x's. Why (b)? Why can no element appear in two distinct rows? Proof: Suppose that hi b = hj x where the x-row was completed after the b-row. Multiply by hj' gives (hj' hi)b = x. This means that x actually appeared in the b th row and so x could not have been chosen since it had already appeared. ------------------------------------------------- Knowing that the order of a subgroup has to divide the order of the group, we can check this out with our familiar Groups S3 and S4. S3 has subgroups of size 1,2,3,6. S4 has subgroups of size 1,2,3,4,6,8,12,24. Remark: Notice that A4 has no subgroup of order 6 even though six divides 12. The order of the subgroup has to divide th order of the group, but it is not true that there is a subgroup for every divisor of the order of the group. A4 = {I, 6 ( , , ), 3 ( , )( , ) } has no subgroup of order six. Proof: Suppose that H were a subgroup of A4 and |H| = 6. We cannot build H out of just the identity and 3-cycles because if we have a 3-cycle, we need its inverse also. This means there are always an even number of 3-cycles which along with the identity makes for an odd number of elements. There are not enough 2-2 cycles to build a group of six elements from them and the identity alone. We thus must have at least one 3-cycle and at least one 2-2-cycles. But any 3-cycle and any 2-2-cycle generate A4 since one can generate 12 rearrangements with them. Remark: If we take a game like Rubics cube, the legal arrangements form a subgroup. This subgroup partitions the set of all rearrangements into identical pieces. For Rubic's cube, there are 12 distinct sets. One way to view this is that if you took the cube apart and put it together randomly, there is only one chance in twelve that you could then solve the cube. Given integers a and b, solving for the coefficients x and y such that ax+by = GCD(a,b). Assignment: (1) Use the subgroup { I, (12) } to partition S3. (2) Use the subgroup 5 Z = {...-10,-5,0,5,10,15 ....} to partition Z. (3) Use the Kleinfour group to partition A4. (4) Use the Kleinfour group to partition S4. Previous Assignment: (1) List all the additive subgroups of Z12. { 0,1,2,3,4,5,6,7,8,9,10,11 } /\ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ { 0,2,4,6,8,10 } { 0,3,6,9 } |\ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \ / | \/ { 0,4,8 } {0,6} \ / \ / \ / \ / \ / \ / V { 0 } (2) List all the additive subgroups of Z19 {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18} | | { 0 } (3) How many additive subgroups are there of Z100. 2^2 5^2 should be 3^2 = 9. 1,2,4,5,10,20,15,50,100 There are 9 divisors of 100. (4) Find integers x and y such that x 13 + y 17 = 1. 1 "Magic Tableau" 13 | 17 1 3 4 13 1 1 4 17 -- 3 0 1 3 13 4 | 13 12 -- 4 answer: 4*13 - 3 17 = 1 1 | 4 4 --- 0