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MTH 561/661 Fall 2018 Abstract Algebra I Drew Armstrong Contents MTH 561/661 Fall 2018 Abstract Algebra I Drew Armstrong Contents Week 1 Quadratic and Cubic Equations, Lagrange's Method 1 Week 2 Permutations, Definition of Groups and Subgroups, Main Examples, Statement of Galois' Theorem 10 Problem Set 1 18 Week 3 Subgroup Generated by a Subset, Intersection and Join of Subgroups, Cyclic Groups, The Circle Group 28 Week 4 Orthogonal and Unitary Groups, Dihedral Groups, Definition of Isomorphism 34 Problem Set 2 42 Week 5 Posets and Lattices, Galois Connections, The Correspondence Theorem, The Fundamental Theorem of Cyclic Groups 52 Week 6 Equivalence Modulo a Subgroup, Lagrange's Theorem, Quotients of Abelian Groups 64 Problem Set 3 74 Week 7 Normal Subgroups, Quotient Groups in General, The First Isomorphism Theorem 81 Week 8 Definition of Automorphisms, Definition of Group Actions, Cayley's Theorem, The Center and Inner Automorphisms 92 Problem Set 4 99 1 Week 9 Direct and Semidirect Products of Groups, Isometries of Euclidean Space 108 Week 10 Simple Groups, The Jordan-H¨olderTheorem, Proof that Sn is not Solvable The Orbit-Stabilizer Theorem, Free and Transitive Actions 119 Problem Set 5 132 Week 11 The Class Equation, Groups of Size p and p2, The Sylow Theorems 142 Week 12 Proof That A5 is Simple, Classification of Finite Simple Groups 148 Problem Set 6 156 Week 1 What is \algebra?" The meaning of the word has changed over time. Here's a historical sketch: In this course we will discuss \modern" (also known as abstract) algebra. But in order to motivate this, I will first talk about pre-modern algebra. Prior to 1830 algebra was understood at the study of polynomial equations. Example. Let a; b; c be any numbers. Find all numbers x such that ax2 + bx + c = 0: I assume that you all learned about quadratic equations in high school. If a = 0 then there is nothing interesting to do, so let us assume that a 6= 0 and divide both sides by a to get b c x2 + x + = 0 a a 2 b c x2 + x = − : a a Now there a famous trick called \completing the square." If we add the quantity (b=2a)2 to both sides then it turns out that the left side factors: b c x2 + x = − a a b b 2 c b 2 x2 + x + = − + a 2a a 2a b b c b2 x + x + = − + 2a 2a a 4a2 b 2 b2 − 4ac x + = : 2a 4a2 Finally, we take \the" square root of both sides and then solve for x: b 2 b2 − 4ac x + = 2a 4a2 p b b2 − 4ac x + = 2a 2a p b b2 − 4ac x = − + 2a p 2a −b + b2 − 4ac = : 2a Wait, I lied. There is no such thing as \the" square root of a number. Actually every number (except 0) has two different square roots. So the \quadratic formula" p −b + b2 − 4ac x = 2a is not really a formula at all, butp more of a \recipe" that tells us how to compute the two roots of the equation. First, let b2 − 4ac denote one of the two square roots of the number 2 b − 4ac. (I don't care whichp one; you can choose your favorite.) Then the other square root is just the negative: − b2 − 4ac. Thus we obtain (in general) two different solutions to the quadratic equation: p p −b + b2 − 4ac −b − b2 − 4ac x = or x = : 2a 2a /// This algorithm was known to ancient civilizations and was slowly put into symbolic form over time. The advantage of the symbolic form is that it encapsulates many different situations and 3 applications. For example: The equation ax2 + bx + c = 0 might represent the intersection of two shapes in the plane; say, a line and a circle. If the quantity b2 − 4ac (called the \discriminant") is negative then its two square roots are imaginary, which means that the line and the circle don't intersect. So far so good, but now we come to a subject that was not known to ancient civilizations. During the Italian Renaissance of the 1500s, a group of mathematicians discovered similar (but more complicated) formulas for the cubic and quartic equations. Example. Let a; b; c; d be any numbers. Find all numbers x such that ax3 + bx2 + cx + d = 0: There is a clever change of variables (never mind the details) that can remove the x2 term. Also, we might as well assume that the leading coefficient is a = 1, otherwise we can just divide both sides by a. Thus it is enough to solve the so-called \depressed cubic:" x2 + px + q = 0: Here's the solution, which is called \Cardano's formula:" s r s r 3 q q2 p3 3 q q2 p3 x = − + + + − − + : 2 4 27 2 4 27 Again, this is not really a formula, but a \recipe" for finding the solutions. Just like \the" squre root, there is no such thing as \the" cube root, since every nonzero number actually has three cube roots. Unfortunately this makes Cardano's formula difficult to interpret, since there might be nine ways to choose the cube roots but the cubic equation x3 − px + q = 0 has only three solutions. We'll discuss a more comprehensive method below. /// The difficulty of interpreting Cardano's formula is the historical reason why people finally accepted the concept of complex numbers. For example, we know that the equation x3 − 15x − 4 = 0 has the real root x = 4, but Cardano's formula gives q p q p x = 3 2 + −121 + 3 2 − −121: p The only wayp to get from thisp expression to x = 4 is to acceptp the fact that 2 + −1 is a cube root of 2 + −121 and 2 − −1 is a cube root of 2 − −121. That is, sometimes the only way to get to a real solution is by going through the imaginary numbers. In modern terms we write C = fa + ib : a; b 2 Rg; p where i = −1 is one of the two square roots of −1. (I don't care which one; you can pick your favorite.) 4 After a burst of activity in the 1500s people got stuck for hundreds of years on the following question. Question. Does there exist a formula for the quintic? Given any numbers a; b; c; d; e; f we consider the equation ax5 + bx4 + cx3 + dx2 + ex + f = 0: Is it possible to write down a recipe for the roots x in terms of the coefficients a; b; c; d; e; f p p p p and the \algebraic" operations +; −; ×; ÷; ; 3 ; 4 ; 5 ;::: ? [Jargon: Is the quintic equation \solvable by radicals?"] In 1770 Joseph-Louis Lagrange wrote an influential book summarizing the state of knowledge of this problem. He had very sophisticated methods that led him to believe that the problem was likely impossible. In 1799 Paolo Ruffini claimed to have a proof of impossibility, but it was flawed. Then in 1823 the young Norwegian mathematician Niels Henrik Abel (1802-1829) finally gave an air-tight proof of the following theorem. The Abel-Ruffini Theorem. Let n ≥ 5. It is impossible in general to write down the roots of an n-th degree polynomial equation in terms of the coefficients and the algebraic operations p p p p +; −; ×; ÷; ; 3 ; 4 ; 5 ;:::: In other words, there exist polynomial equations of all degrees ≥ 5 that are not solvable by radicals. This was a historic achievement but Abel's proof was long and complicated. Furthermore, there are certain special kinds of polynomials that can be solved by radicals. Example. The quintic equation ax5 + b = 0 is easily solvable. Here's the recipe: ax5 + b = 0 ax5 = −b b x5 = − a r b x = 5 − : a Now one just has to compute the five 5th roots of the number −b=a. [Remark: How to do this? In general one can use numerical methods (like Newton's method) 5 to find one particular fifth root, say α 2 C where α = −b=a. Then the complete list of 5th roots is α; !α; !2α; !3α; !4α; 5 where ! = e2πi=5 is called a \primitive 5th root of 1."] That was sort of obvious, but there are more complicated examples. Example. De Moivre's quintic has the form 1 x5 + ax3 + a2x + b = 0: 5 It turns out (you don't want to see the details) that the roots of this equation r1; r2; r3; r4; r5 are given by the formula j −j rj = ! u1 + ! u2; 2πi=5 where ! is any primitive 5th root of 1 (e.g. ! = e ) and where u1 and u2 are any two numbers satisfying ( 5 5 u1 + u2 = −b; 5 5 5 u1 · u2 = (−a=5) : 5 5 [Exercise: Use the quadratic formula to solve for the numbers u1 and u2.] Long story short: De Moivre's quintics are solvable by radicals. So the next order of business was to clean up the details and understand the distinction between solvable and non-solvable polynomials. Abel intended to do this. Abel's Research Program. Find a method to determine which polynomials are solvable by radicals, and which are not. But then he died suddenly (of tuerculosis) at the age of 26. This brings us to the year 1829. At this time, a young frenchman named Evariste´ Galois was working in obscurity on similar problems. Galois had some brilliant and visionary ideas, but he also had a volatile personality and he died in a dual in 1832 at the age of 20, before he could gain any recognition for these ideas.
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