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Algebra. A graduate course by Isaacs I.M.

By Isaacs I.M.

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R < n. Then xm = e becomes PROOF. xqn+r=e xqnx'=e (xn)qx'=e. But X n= e, so the last equation becomes x'=e. But r is smaller than n, so this is impossible unless r = 0. Thus m= qn + 0, so n divides m. iii) Here we use our information about greatest common divisors. We must show that n/ d is the smallest positive integer k such that (xml =e. First of all, (xmf/d = xm·(n/d)= xO is smaller than n/d and (xm)k=e. We will show that n/ d divides k, a contradiction.

The elements of this group are n-tuples ( g 1,g2 , ••• ,gn) with g; E G;, and the multiplication is defined componentwise. As a matter of fact, there is no reason why we have to restrict ourselves to finitely many factors, but we will rarely use infinitely many. Examples 1. Let G1 = G2 = · · · = Gn = (R, + ). Then is ordinary n-space Rn under addition of n-tuples. 2. Consider 7L 2 X 7L 2, where the operation on each factor is addition mod 2. This is a group of order 4, and already reveals some interesting things about direct products.

3)? (4)? (5)? (8)? 2 Let G be the group of all real-valued functions on the real line under addition of functions, and letfEG be the function such thatf(x)=l for all xEIR. Indicate what sort of configuration you would get if you drew the graphs of all the functions in (f) on one set of axes. 3 Let X={1,2,3,4,5}. ), how many elements are there in (A)? What are they? Section 4. 4 In (Z: 30, EB), find the orders of the elements 3, 4, 6, 7, and 18. 5 Let G be a group and let x E G be an element of order 18.

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