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 ############################################## # Test 2 Friday October 19 # # over material since last test # ############################################## Friday, October 12. Main idea: When you point a finger at someone there are three other fingers pointing back at you. Key words: First homomorphism theorem, isomorphism Goal: Learn and understand the first homomorphism theorem. Previous Assignment: 1. Let H = {I,(12)} which is a subgrout of S3. (a) Show that H is NOT normal. (13) {I, (12) } (13) = {I, (23)} =/= H (b) Compute H(123) and H(132). {I,(12)} (123) = {(123),(23)} {I,(12)} (132) = {(132),(13)} (c) Compute the product H(123) H(132) {(123),(23)}{(132),(13)} = {I,(123)(13),(23)(132),(23)(13)} = {I,(23),(12),(123)} which is not a coset at all. (d) What elements of H(123) H(132) are not in He. {(23),(123)}. 2. K4 is a normal subgroup of S4. (a) Create the elements in the cosets of S4/K4. _________________________________________ | | | (132) (143) (234) (124) | (132) | | |_________________________________________| | | | (123) (243) (142) (134) | (123) | | |_________________________________________| | | | (23) (1243) (1342) (14) | (23) | | |_________________________________________| | | | (13) (1432) (24) (1234) | (13) | | |_________________________________________| | | | (12) (34) (1423) (1324) | (12) | | |_________________________________________| | | | I (12)(34) (13)(24) (14)(23) | | | |_________________________________________| (b) Find the orders of each of the elements of S4/K4. The top two rows are of order 3, the bottom is of order 1, the remaining 3 are of order 2. 3. G = Z4 x Z6 is a commutative group with 24 elements. H = <(2,3)> is the subgroup generated by (2,3). (a) List the twelve cosets of G/H. _____________________________________ | (3,5), (1,2) | (3,5) |___________________________________| | (2,4), (0,1) | (2,4) |___________________________________| | (1,3), (3,0) | (1,3) |___________________________________| | (4,2), (2,5) | (0,2) |___________________________________| | (3,1), (1,4) | (3,1) |___________________________________| | (2,0), (0,3) | (2,0) |___________________________________| | (1,5), (3,2) | (1,5) |___________________________________| | (0,4), (2,1) | (0,4) |___________________________________| | (3,3), (1,0) | (3,3) |___________________________________| | (2,2), (0,5) | (2,2) |___________________________________| | (1,1), (3,4) | (1,1) |___________________________________| | (0,0), (2,3) | (0,0) |___________________________________| (b) From the commutative group theory G/H = Z4xZ3 or G/H = Z2xZ2xZ3. Explain which it is and why you think so. It is cyclic so must be Z4xZ3. Theorem: A homomorphism is 1-1 if and only if Ker(f) = {e}. We already know that the preimages are the cosets. For each of these cosets to have exactly one element is equivalent to having the kernel have exactly one element. Since the identity e is automatically in the kernel, the kernel must be {e}. Fundamental Homomorphism Theorem Theorem(3.2.10 page 176). Let H be a normal subgroup of G. Then pi:G-->G/H given by pi(g) = gH is a homomorphism with kernel H. Proof: We already know that when H is a normal subgroup, The set of cosets G/H is a group under the product Ha*Hb = H(ab). The only thing about this that was difficult was to prove that the product is well defined. Certainly pi(ab) = H(ab) = Ha*Hb = pi(a)*pi(b) so pi is a homomorphism. The kernel of pi is the set of all elements which map to He. Since pi maps the elements to the cosets which contain them, the elements which map to the coset He are the elements of H. This means Ker pi = H. Theorem(3.2.12 page 176) Let f:G-->W be a group homomorphism with kernel K. Then G/K is isomorphic to the image of f. f G----------> Image[G] c W | ___ | /| | / | | / | / | / | / t | / | / | / \|// |/ G/Ker Define t(Kg) = f(g). Notice that this is well defined since everything in the coset Kg maps to the same element. The rest is easy. t(Ka*Kb) = t(Kab) = f(ab) = f(a)f(b) = t(Ka)t(Kb) so t is a homomorphism. The kernel of t are the cosets which map to eW. But t(Ka) = eW implies f(a) = eW and so a is K and Ka = K. Since the kernel is a singleton, the whole map t is 1-1. and the map is onto the image by construction. If f(a) is in the image, the t(Ha) = f(a). Note: A homomorphism which is 1-1 and onto is called an isomorphism.