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 Wednesday, September 19, 2001 ********************************** * TEST MONDAY SEPTEMBER 24 * * pages 51 through 135 * * pages 197 through 208. * ********************************** The material we are covering today is not in the book at all, at least as far as I can tell. It is the nuts and bolts of the Fundamental Theorem of Finitely Generated Abelian Groups that is stated on page 133. Main Idea: An efficient method find the cyclic subgroup types. Key Words: Finitely Generated, Generators, Relations. Objective: Be able to construct the cyclic subgroups into which a finitely generated commutative group is composed. Determine the structure of the commutative group defined by generators a,b,c and relations 3a + 9b + 9c = 0 and 9a - 3b + 9c = 0. First, with out any constraints, the commutative group generated by a,b,c will be Z x Z x Z. Or internally x x , The elements can be represented as n1 a + n2 b + n3 c or (n1,n2,n3) where n1,n2,n3 are integers. Certainly this group decomposes into the direct sum of cyclic groups, namely as Z x Z x Z. However when there are constraints involved, we still have an abelian group, but now it will be isomorphic to Zp x Zq x Zr. So we want to find numbers p,q,r. We also want to recognize the generators internally. The method is given here: It is very much like row reduction in matrices, except that we add multiplies of rows to other rows, and we add multiples of columns to other columns. We start with the matrix of the constraints and we want to finish with a diagonal matrix of the form r 0 0 0 0 0 0 s 0 0 0 0 0 0 t 0 0 0 0 0 0 0 0 0 Then the answer is Zr x Zs x Zt x Z x Z x Z In our particular case: 3 9 9 9 -3 9 add -3xrow 1 to row 2 3 9 9 0 -30 -18 add -3 x column 1 to column 2 and column 3 3 0 0 0 -30 -18 add -2 x column 3 to column 2 3 0 0 0 6 -18 add 3 x column 2 to column 3 3 0 0 0 6 0 The answer is Z3 x Z6 x Z. There is an element a = (i1,i2,i3) and an element b = (j1,j2,j3) and an element c = (k1,k2,k3) such that 3a+9b+9c = 0 and 9a-3b+9c = 0 We find a,b,c by doing the column operations backwards. In our particular case: a b c (* Put a,b,c atop the columns*) 3 9 9 9 -3 9 add -3xrow 1 to row 2 (* row operations do not affect column tops *) a b c 3 9 9 0 -30 -18 add -3 x column 1 to column 2 and column 3 (* Do the opposite to the column tops. *) a+3b+3c b c (* Add 3 times column 2 and 3 times column*) 3 0 0 (* 3 to column 1. *) 0 -30 -18 add -2 x column 3 to column 2 (* Do the opposite to the column tops. *) a+3b+3c b 2b+c (* Add 2 times column 2 to column 3. *) 3 0 0 0 6 -18 add 3 x column 2 to column 3 (* Do the opposite to the column tops. *) a+3b+3c -5b-3c 2b+c (* Add -3 times column 3 to column 2. *) 3 0 0 0 6 0 The generators are a' = a+3b+3c b' = -5-3c c' = 2b+c. 3a' = 0 a' generates Z3 6b' = 0 b' generates Z6 c' has infinite order. c' generates Z. | a' | | 1 3 3 || a | | b' | = | 0 -5 -3 || b | | c' | | 0 2 1 || c | | a | | 1 3 6 || a'| | b | = | 0 1 3 || b'| | c | | 0 -2 -5 || c'| In Z3 x Z6 x Z a = (1, 3, 6) b = (0, 1, 3) c = (0,-2,-5) 3a+9b+9c = (3,-6, 0) = (0,0,0) 9a-3b+9c = (9, 6, 0) = (0,0,0) The overall method of making these matrix reductions is: (1) Find the smallest element (s) in the matrix and move it to the top left hand corner. If (s) is negative, make it positive by multiplying the row of (s) by -1. (s) * * * * . . . * . . . * . . . * . . . Now you can subtract off multiples of s from both the row of (s) and the column of (s). This way, you either get entries in the row and column of (s) to be zero, or numbers less than (s). If they are not all zeroed out, put the new smallest one in the top left and continue. Since you cannot get smaller for- ever, eventually you will zero out the row and column of (s). Then continue with the second row and second column position. Assignment: 1. Given the generators and relations 3a + 6 b + 12 c = 0 8a + 2 b + 6 c = 0 2a + 3 b + 15 c = 0 Decompose the group into cyclic summands and give the generators of each cyclic summand. 2. Given the generators and relations 8a + 5 b + 12 c = 0 6a + 3 b + 2 c = 0 6a + 9 b + 6 c = 0 Decompose the group into cyclic summands and give the generators of each cyclic summand. 3. Given the generators and relations 12a + 9 b + 2 c = 0 8a + 3 b + 4 c = 0 4a + 4 b + 6 c = 0 Decompose the group into cyclic summands and give the generators of each cyclic summand.