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: Normal subgroups split the group into cohesive bunches. Key words: Factor Group, G/H Goal: Learn about the Group Structure of the Cosets. Previous Assignment: (1) D: polynomials -> polynomials where D is differentiation. What is the Kernel of D? ans: The kernel of D consists of all polynomials which map to zero under differentiation. These are the constant polynomials. (2) If G is a commutative group, show that the map f(x) = x^2 is a homomorphism. Proof: f(xy) = (xy)^2 = xyxy = xxyy = x^2y^2 = f(x)f(y) (3) In Z # = {The elements relatively prime to 36 under multiplication} 36 Find the kernel of the map f(x) = x^2. The Kernel of f is {1,17,19,35} (4) Let f:G-->H be a group homomorphism. (a) Show that ker(f) is a subgroup of G. (b) Show that g ker(f) g' = ker(f) for all g in G. (i.e. Show that ker(f) is a normal subgroup of G.) Part(a) Closure: If g1,g2 are in Kernel of f, then f(g1*g2) = f(g1)of(g2) = eH o eH = eH. Therefore g1*g2 is also in the kernel of f. Inverses: If g is in the kernel of G then f(g') = f(g)' = (eH)' = eH. Therefore g' is also in the kernel of f. Part(b). If x is in kernel of f and g is in G, then f(gxg') = f(g) f(x) f(g') = f(g) eH f(g') = f(g) f(g') = f(gg') = f(e) = eH Therefore gxg' is in the kernel of f. We have shown that g ker(f) g' c kerf(f). Therefore ker(f) is normal. Before we take a very short break from homomorphisms to look at normal subgroups themselves, we wish to better understand the relationship of the coset containing an element g with the image f(g) under a homomorphism f. Take any function f:S-->T, where S and T are just sets and f is any function. We can collect the elements of S into cells by gathering together all the elements which are mapped to the same element. _____________________________ ________________ | | | | | Elements mapping to t1 | | t1 | |---------------------------| | | | | | t2 | | Elements mapping to t2 | | | |---------------------------| | | | | | t3 | | Elements mapping to t3 | | | |---------------------------| | . | | : | | . | | : | | . | |___________________________| |______________| This collection of cells (subsets) is called a partition. Subsets of S are called a partition of S if: (a) They are disjoint. (No element is in two different cells). (b) Taken together they contain all the elements of S. (c) None of the cells are empty. If f is a group homomorphism from G to H, then the cells are simply the cosets of G using the subgroup ker(f). The following is given in the book on page 166. Theorem 3.1.15 Let f:G->H be a group homomorphism. Let K = Ker(f). Then the preimages of elements of H are just the cosets {gK} Proof: (i) Every element of gK and Kg is mapped to the same element by f. Suppose x is in K. f(gx) = f(g)f(x) = f(g) eH = f(g). f(xg) = f(x)f(g) = eH f(g) = f(g). (ii) Every element mapped to f(g) is in Kg and gK. Suppose f(q) = f(g). then f(q)f(g)' = eH f(qg') = eH so qg' is in K. then q = qg' g is in Kg. Similarly f(g'q) = eH so g'q is in K. And q = g g'q is in gK. So far we have shown that the elements of Kg and gK map to f(g) and these are the only elements which map to f(g). Probably we should point out that Kg=gK since K is normal. Now we forget about homomorphism and just consider a group G with a normal subgroup H. Then (a) Ha = aH (b) (Ha)(Hb) c H(ab) Proof of (a). This is true because H is normal. Given any a in G, aHa' = H so for any h1 in H, a h1 a' = h2 Thus a h1 = h2 a so aH c Ha. Also a' H (a')' = H so for any h1 in H a' h1 a = h2 h1 a = a h2. This shows Ha c aH. Since the containment is both ways, we have that aH = Ha. Proof of (b). We claim that any element in Ha times any element in Hb will be some element in H(ab). if h1 a is in Ha and h2 b is in Hb, then (h1 a)(h2 b) = h1 (a h2) b = h1 (h3 a) b = (h1 h3)(ab) c H(ab). (since a h2 = h3 a for some h3) This is a very powerful theorem. It says that There is a (well defined) product on the cosets of G/H. define this product by (Ha)(Hb) = H(ab). This product is associative, has an identity, and has inverses. Associativity: (Ha Hb) Hc = H(ab) Hc = H (ab)c Ha (Hb Hc) = Ha H(bc) = H a(bc) Since (ab)c = a(bc) are equal in the group G, the coset products are equal. Identity: Ha He = Hae = Ha He Ha = Hea = Ha Inverse: Ha Ha' = H(aa') = He Ha' Ha = H(a'a) = He The cosets form a group under coset multiplication. =================================================== The notation for the cosets generated by any subgroup H is indicated by G/H. Note that this is not really well defined because one gets a different partition depending if you build the cosets on the right side or on the left side. But one assumes that it is somehow understood which side is meant. Only when H is normal does the coset product make sense. It requires H being normal before Ha Hb c H(ab). When H is normal, then G/H is also a group and it is called the quotient group (of G by the normal subgroup H). Example: In the integers, the multiples of n = nZ = {... -4n, -3n -2n -n, 0, n,2n,3n,4n, ....} form a subgroup. It is normal because the integers are commutative. The quotient group Z/nZ is Zn. This shows that the integers mod n are associative. This is something left till now because it is very time consuming to prove Zn is associative from the "use the remainder" rule. Assignment: 1. Let H = {I,(12)} which is a subgrout of S3. (a) Show that H is NOT normal. (b) Compute H(123) and H(132). (c) Compute the product H(123) H(132) (d) What elements of H(123) H(132) are not in He. 2. K4 is a normal subgroup of S4. (a) Create the elements in the cosets of S4/K4. (b) Find the orders of each of the elements of S4/K4. 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. (b) From the commutative group theory G/H = Z4xZ3 or G/H = Z2xZ2XZ3. Explain which it is and why you think so.