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, October 10. Main idea: The operation is preserved. Key words: Homomorphism Goal: Learn the basic properties of homomorphisms. Previous Assignment: The corresponding numbering for triangular puzzle peg is p 1 q q p 1 p 1 q p 1 q p 1 q q p 1 q p 1 p 1 q p 1 q p 1 q p 1 q p 1 q (1) For this puzzle, remove one peg and jump down to one peg. Which peg do you remove and where is the last peg? o o o o o o o o o o 1 1 1 2 o 3 2 2 3 3 The 1's, 2's, and 3's cancel. Thus the parity of this board is the same as the center, which is also the same as the three corners. If the board is to be solved, we cannot remove a corner peg nor the center peg because then the parity of the board would be zero and be unsolvable. If we remove an edge peg, the parity of the board will be the same as the edge peg adjacent to the peg removed. The last peg will be left next to the original hole, or the similar symmetrical places on the other two edges. (2) Is this puzzle possible or impossible to solve if you are allowed to jump outside the figure? Why? . . . . o . . o o . . o o o . . o o o o . . o o o o o . . . . . . . . . . . . . 1 . . 1 1 . . 2 3 4 . . 2 3 4 5 . . 2 3 4 5 5 . . . . . . . . . The like numbered positions cancel. Thus the parity of the game is zero. It cannot be reduced to one peg. (3) Prove that the Kleinfour group is a normal subgroup of S4. Use the fact that that the Kleinfour group contains all 2-2 cycles of S4. K4 consists of the identity I and all of the 2-2-cycles. gIg' = I so any conjugate of the identity is back in K4. If x is any other element in K4, x is a 2-2-cycle. Since gxg' is again a 2-2 cycle, gxg' is in K4. Thus gK4g' c K4 and K4 is a normal subgroup. (4) (a) Find an even permutation pi such that pi (12)(34) pi' = (14)(23) pi = (243) <== ans (b) Find an odd permutation pi such that pi (12)(34) pi' = (14)(23) First, write the same permutation in an equivalent way. pi (12)(34) pi' = (14)(32) pi = (24) <=== ans (5) Show that the subset of S5 containing the identity and all 2-2 cycles is not closed. (12)(34) (12)(35) = (1)(2)(354) = (354) This is a 3-cycle. Abstract algebra can be used to prove something is impossible. This is done with the observation of something called an "invariant." Something is an invariant if it is not changed, or if it is changed in some predictable way. The algebraic argument goes like this. Statement: It is impossible to to such and such. Why? Because to do so would involving changing an invariant. Therefore it must be impossible to do that. Examples 1: The 0,1,p,q invariants of Puzzle Peg. The argument is: We cannot solve that puzzle because the parity at the start is 0. The parity must remain zero throughout the entire process and will be still zero at the end. It is impossible to have one peg and be zero parity. Therefore you cannot jump the board down to one peg. Example 2: The 16-puzzle. The argument goes, if we could change position A to B, then it would require an even number of moves. But the change from A to B is an odd permutation. Therefore the change is impossible. Example 3: It is impossible to trisect 60 degrees. There is a property that holds for every point that is constructed. If we could trisect 60 degrees we could construct a point without that property. Therefore we cannot trisect 60 degrees. Example 4: It is impossible to write a quintic formula. We have a quadratic formula, we have a cubic formula, we have a quartic formula. But there is no formula for a degree 5 polynomial. The invariant in this case is the existence of normal subgroups in Sn. Since these normal subgroups do not exist for S5, the the quintic formula cannot exist either. The material is found on pages 161-169 of section 3.1 of chapter 3 starting with "Homomorphisms and Factor Groups." Definition: Let (G,*) and (H,o) be groups. A function f:G-->H is called a homomorphism if f(g1*g2) = f(g1)og(g2) for all g1,g2 in G. We see this behavior so often that we often expect it. D[f+g] = D[f]+D[g]. INT[f+g] = INT[f] + INT[g]. x+y x y e = e * e Log[ab] = Log[a]+Log[b]. There are a lot of properties that follow directly. Theorem 3.1.12: If (G,*) and (H,o) are groups and f:G-->H is a homomorphism, then (a) f maps the identity of G to the identity of H. (b) f preserves inverses. I.e f(g') = f(g)' for all g in G. (c) f maps subgroups of G to subgroups of H. -1 (d) f maps subgroups of H to subgroups of G. Proof of (a). f(e ) = f(e *e ) = f(e )of(e ). Multiply ------------ G G G G G by the inverse of f(e ) in H gives e = f(e ). G H H Proof of (b). e = f(e ) = f(g*g') = f(g)of(g') ------------ H G e = f(e ) = f(g'*g) = f(g')of(g) H G Since f(g')of(g) = f(g)of(g') = e , by definition of H the inverse of f(g) in H, f(g)' = f(g'). Proof of (c). Let A be a subgroup of G. ----------- f[A] = {f(a) | a is in A}. Closed. f(a )of(a ) = f(a *a ) is in f[A]. 1 2 1 2 Inverses. f(a)' = f(a') is in f[A]. Proof of (d). Let B be a subgroup of H. ----------- -1 f [B] = {g in G such that f[g] is in B}. Closed. If f[g ] is in B and f[g ] is in B, then 1 2 f[g *g ] = f[g ]of[g ] is in BoB c B. 1 2 1 2 -1 Thus g *g is in f [B]. 1 2 Inverses. If f[g] is in B, then f[g'] = f[g]' is in B, -1 since B contains inverses, Thus g' is in f [B]. Page 166 Definition of the kernel. Definition 3.1.13: Let f:G--H be a homomorphism of groups. The subgroup Ker(f) = {g in G | f(g) = e } is called the kernel of f. H Assignment: (1) D:polynomials -> polynomials where D is differentiation. What is the Kernel of D. (2) If G is a commutative group, show that the map f(x) = x^2 is a homomorphism. (3) In Z # = {The elements relatively prime to 36 under multiplication} 36 find the kernel of the map f(x) = x^2. (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.)