Enrichment - Enrichment Chapters 1–4 form an idealized...

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Unformatted text preview: Enrichment Chapters 1–4 form an idealized undergraduate course, written in the style of a graduate text. To help those seeing abstract algebra for the first time, I have prepared this section, which contains advice, explanations and additional examples for each section in the first four chapters. Section 1.1 When we say that the rational numbers form a group under addition, we mean that rational numbers can be added and subtracted, and the result will inevitably be rational. Similarly for the integers, the real numbers, and the complex numbers. But the integers (even the nonzero integers) do not form a group under multiplication. If a is an integer other than ± 1, there is no integer b such that ab = 1. The nonzero rational numbers form a group under multiplication, as do the nonzero reals and the nonzero complex numbers. Not only can we add and subtract rationals, we can multiply and divide them (if the divisor is nonzero). The rational, reals and complex numbers are examples of fields , which will be studied systematically in Chapter 3. Here is what the generalized associative law is saying. To compute the product of the elements a,b,c,d and e , one way is to first compute bc , then ( bc ) d , then a (( bc ) d ), and finally [ a (( bc ) d ) e ]. Another way is ( ab ), then ( cd ), then ( ab )( cd ), and finally ([( ab )( cd )] e ). All procedures give the same result, which can therefore be written as abcde . Notice that the solution to Problem 6 indicates how to construct a formal proof of 1.1.4. Section 1.2 Groups whose descriptions differ may turn out to be isomorphic, and we already have an example from the groups discussed in this section. Consider the dihedral group D 6 , with elements I , R , R 2 , F , RF , R 2 F . Let S 3 be the group of all permutations of { 1 , 2 , 3 } . We claim that D 6 and S 3 are isomorphic. This can be seen geometrically if we view D 6 as a group of permutations of the vertices of an equilateral triangle. Since D 6 has 6 elements and there are exactly 6 permutations of 3 symbols, we must conclude that D 6 and S 3 are essentially the same. To display an isomorphism explicitly, let R correspond to the 1 2 permutation (1,2,3) and F to (2,3). Then I = (1) , R = (1 , 2 , 3) , R 2 = (1 , 3 , 2) , F = (2 , 3) , RF = (1 , 2) , R 2 F = (1 , 3) . If G is a nonabelian group, then it must have an element of order 3 or more. (For example, S 3 has two elements of order 3.) In other words, if every element of G has order 1 or 2, then G is abelian. To prove this, let a,b ∈ G ; we will show that ab = ba . We can assume with loss of generality that a 6 = 1 and b 6 = 1. But then a 2 = 1 and b 2 = 1, so that a is its own inverse, and similarly for b . If ab has order 1, then ab = 1, so a and b are inverses of each other. By uniqueness of inverses, a = b , hence ab = ba . If ab has order 2, then abab = 1, so ab = b − 1 a − 1 = ba ....
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This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.

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Enrichment - Enrichment Chapters 1–4 form an idealized...

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