{ "cells": [ { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "\n", "\n", "
\n", "\n", "\n", "\n", "\n", "\n", "Abstract Algebra: An Interactive Approach, 2e\n", "
\n", "\n", "\n", "©2015 This notebook is provided with the textbook, "Abstract Algebra: An Interactive Approach, 2nd Ed." by William Paulsen. Users of this notebook are encouraged to buy the textbook.\n", "
\n", "\n", "\n", "\n", "\n",
"Chapter 1
\n",
"Understanding the Group Concept\n",
"
Initialization: This cell MUST be evaluated first:
" ] }, { "cell_type": "code", "execution_count": 0, "metadata": { "collapsed": false }, "outputs": [ ], "source": [ "%display latex\n", "try:\n", " load('absalgtext.sage')\n", "except IOError:\n", " load('/media/sf_sage/absalgtext.sage')" ] }, { "cell_type": "markdown", "metadata": { "collapsed": false }, "source": [ "Stay · x = x and x · Stay = x
\r\n", "for all dance steps x. We call Stay the identity of the dance steps.x·y is sometimes not equal to y·x.
\r\n", "This is perhaps the biggest difference between Terry's algebra and the algebra that we are familiar with. We are familar with the \r\n", "commutative property of numbers:\r\n", "x y = y x for all real numbers x and y.
\r\n", "Since Terry's dance steps do not obey this law, we say that the dance steps are non-commutative.x (y z) = (x y) z for all numbers x, y, and z.
\r\n", "This is called the associative law of multiplication. Do Terry's dance steps obey this law?x·y = Stay.
\r\n", "The dance step y is called an inverse of x. For example, RotLft is the inverse of RotRt, and \r\n", "Spin is the inverse of Spin. \r\n", "\r\n", "(x·y)-1 = y-1·x-1.
\r\n", "Can you prove this? \r\n", "\r\n", "\r\n", " \r\n", "PROPOSITION 1.2(x·y)-1 = y-1·x-1.
\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "\r\n", "Let us see if we can find any more patterns in Terry's multiplication table. Here is the table again:Spin · x = RotRt.
\r\n", "So we can say that each row contains exactly one of every element by the following statement:\r\n", "\r\n", "\r\n", " \r\n", "PROPOSITION 1.3a·x = b.
\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "\r\n", "Can you use this proposition to show that any column of the table contains only one of each element? This is essentually Problem 6.x ≡ y (mod n)
\r\n", "if, and only if, there is an integer k such that \r\n", "(x − y) = k n.
\r\n", "\r\n", "Note the slight difference in notation between the operator mod (expressed in boldface) and the above notation (where mod is not in boldface). The two notations are clearly related, since x ≡ y (mod n) means that x mod n = y mod n.x ≡ y (mod n)
\r\n", "forms an equivalence relation on the set of integers.[a] = { s ∈ S | s ∼ a}.
\r\n", "\r\n", "EXAMPLE:[3] = {… -37, -27, -17, -7, 3, 13, 23, 33, 43, …}.
\r\n", "Other equivalence classes in this relation are similar.x ≡ a (mod n) and y ≡ b (mod n),
\r\n", "then x + y ≡ a + b (mod n) and x y ≡ a b (mod n).4 + 5 ≡ 2 (mod 7), 4 × 5 ≡ 6 (mod 7).
\r\n", "\r\n", "2 x = 1 + 10 n
\r\n", "for some n.e·x = x·e = x.
\r\n", "3) (inverse) For every x in G, there exists a y in G, called the inverse of x, such that \r\n", "x·y = e.(a ·b)·c = a·(b·c).
\r\n", "x ∈ G.
\r\n", "\r\n", "Thus, the symbol ∈ means "is an element of."{0, 1, 2, …, n − 1},
\r\n", "with the opperator (·) being the sum modulo n. This group will be denoted Zn. For example, \r\n", "the group Z10 is given by:a + b = b + a for all a and b in ℤ.
\r\n", " \r\n", "EXAMPLE:f(x) = m x + b, with m, b ∈ ℝ, m ≠ 0.
\r\n", "(The fancy ℝ here represents the real numbers.) We multiply two linear functions together by function composition. That is, if \r\n", "f(x) = m x + b, and g(x) = n x + c, then\r\n", "f·g = f(g(x)) = m(n x + c) + b = \r\n", "(m n) x + (m c + b)
\r\n", "which is again a linear function. Note that in f·g, we do g first, then f, so we apply the functions from right to left. We \r\n", "can find the inverse of f(x) as well:\r\n", "f -1(x) = 1⁄m x − \r\n", "b⁄m,
\r\n", "which is also a linear function. The identity function is just e(x) = x. The associativity property of function composition completes \r\n", "the requirements for the set of all linear functions to constitute a group.(e·e)·e = e·(e·e) = e.
\r\n", "We can give a multiplication table of this group by executing the following:x, | \r\n", "
x·x, | \r\n", "
x·x·x, | \r\n", "
x·x·x·x, and | \r\n", "
x·x·x·x·x. | \r\n", "
x = | \r\n", "x1, | \r\n", "
x·x = | \r\n", "x2, | \r\n", "
x·x·x = | \r\n", "x3, etc. | \r\n", "
xn = xn−1·x.
\r\n", "By defining the nth power in terms of the (n − 1)st power, we inductively have defined xn \r\n", "whenever n is a positive integer.x−n = (xn)-1 if n > 0.
\r\n", "Note that if the operator · represents addition, the raising x to a power is repeated addition, which is multiplication.xm+n = xn·xm.
\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "\r\n", "This last proof utilizes an important method of proving theorems called induction, which was introduced in §0.1. Induction is based on the well \r\n", "ordering axiom, which states that any non-empty subset of positive integers contains a smallest element.(xm)n = x(m n)
\r\n", "\r\n", "Click here for the proof.\r\n", "\r\n", "\r\n", "Propositions 1.8 and 1.9 show that the common laws of exponents hold for elements of a group. In the next section, we will use the powers of elements to explore the properties of a group.\r\n", "\r\n", "x·(y·z) = (x·y)·z.
\r\n", "On the left side, we see that y·z is an identity element, so x·(y·z) = x. \r\n", "But on the right side, we find that x·y is an identity element, so (x·y)·z = z. Thus, \r\n", "x = z, and so x is an inverse of y. Therefore, the inverse of an inverse gives us back the original element.(x·y)·(y-1·x-1) = \r\n", "x·(y·y-1)·x-1 = x·Stay·x-1 = \r\n", "x·x-1 = Stay.
\r\n", "So (x·y)-1 = y-1·x-1.a-1·(a·x) = a-1·b.
\r\n", "Then \r\n", "(a-1·a)·x = a-1·b.
\r\n", "Stay·x = a-1·b.
\r\n", "So x = a-1·b. Thus, if there is a solution, this must be the unique solution a-1·b. Let \r\n", "us check that this is indeed a solution.\r\n", "a·x = a·a-1·b = Stay·b = \r\n", "b.
\r\n", "Thus, there is only one solution to the equation, namely a-1·b.x − z = (x − y) + (y − z) = \r\n", "k1 n + k2 n = (k1 + k2)n.
\r\n", " \r\n", "Hence, we find that x ≡ z (mod n).x·y ≡ 1 (mod n).
\r\n", "But this means that x y = 1 + w n for some w. This is impossible, since x y is a multiple of p, but 1 + w n is one \r\n", "more than a multiple of p.u x + v n = gcd(x, n) = 1.
\r\n", "But then \r\n", "u x = 1 + (−v) n,
\r\n", "and so u·x ≡ 1 (mod n). Hence, u is a multiplicative inverse of x.xm+0 = xm = xm·e = \r\n", "xm·x0, x0+n = xn = \r\n", "e·xn = x0·xn.
\r\n", "We will now prove the statement when m and n are positive integers. If n is 1, then we have\r\n", "xm+1 = x(m+1)−1·x = \r\n", "xm·x1,
\r\n", "using the inductive definition of the power of x.xm+(k−1) = xm·xk−1.
\r\n", "But then\r\n", "xm+k = xm+k−1·x = \r\n", "xm·xk−1·x = xm·xk.
\r\n", "Thus, by assuming the statement is true for n = k − 1, we found that it was also true for n = k. By induction, \r\n", "this proves that xm+n = xm·xn for all positive n.(xm+n)-1 = \r\n", "(xn)-1·(xm)-1.
\r\n", "But by the definition of negative exponents, this is\r\n", "x(−n)+(−m) = \r\n", "x−n·x−m
\r\n", "which, by letting M = −n and N = −m, proves the proposition for the case of both exponents being negative.xm = x(m+n)+(−n) = \r\n", "xm+n·x−n.
\r\n", "So we have xm·(x−n)-1 = \r\n", "xm+n·x−n·(x−n)-1, and hence\r\n", "xm+n = xm·xn.xn = x(−m)+(m+n) = \r\n", "x−m·xm+n.
\r\n", "So we have (x−m)-1·xn = \r\n", "(x−m)-1·x−m·xm+n, and \r\n", "hence xm+n = xm·xn.(xm)0 = e = xm·0, \r\n", "(xm)1 = xm = xm·1
\r\n", "We will again proceed by means of induction, which means we can assume that the statement is true for the previous case, with n replaced by n − 1.\r\n", "That is, we can assume that\r\n", "(xm)n−1 = xm(n−1).
\r\n", "Note that\r\n", "(xm)n = \r\n", "(xm)n−1·xm = \r\n", "xm(n−1)·xm.
\r\n", "By Proposition 1.8, this is equal to x(m(n−1)+m) = xm n.(xm)n = ((xm)−n)-1 = \r\n", "(x−m n)-1 = xm n.
\r\n", "If n is negative, then −n is positive, so the second step is valid.