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Unformatted text preview: Chapter 6 Finite Abelian Groups In this chapter, we shall give a complete classification of finite abelian groups. In so doing, we shall observe how the assumption that our binary operation is commutative brings considerable restriction and so makes abelian groups very tractible to study. We shall also observe how useful direct products are in our classification. Before we describe the structure of finite abelian groups, we make a number of comments. The first is that when studying abelian groups it is common to use additive notation for the binary operation rather than mul tiplicative. The reason for this is that the cyclic groups Z of all integers and Z n of integers modulo n are very important. Many textbooks make this change in notation and some previous exam papers have followed this con vention. The author of these lecture notes has chosen, however, to continue to use multiplicative notation so as to maintain consistency with other chap ters in the lecture course and not to introduce a whole new set of notation for everything we have covered so far. We recall some facts that we need: (i) In an abelian group, every subgroup is normal (Example 3.3). (ii) As a consequence, if H and K are subgroups of an abelian group, then HK = { hk  h H, k K } is also a subgroup (Problem Sheet III, Question 3). (iii) If x and y are elements of a finite abelian group, then x and y commute, so o ( xy ) = lcm ( o ( x ) ,o ( y ) ) (Problem Sheet II, Question 5). In our classification of finite abelian groups, the constituents will turn out to be direct products and cyclic groups. We start by proving: 58 Lemma 6.1 Let m and n be coprime positive integers. Then C mn = C m C n . (Recall that m and n are coprime if the only positive common divisor is 1. Also C m continues to denote a multiplicatively written cyclic group of order m .) Proof: Let C m = ( x ) and C n = ( y ) . We know that the direct product C m C n is a group of order mn . Consider g = ( x,y ) C m C n . The k th power is given by g k = ( x k ,y k ) and this equals the identity if and only if x k = 1 and y k = 1; that is, when m = o ( x ) and n = o ( y ) both divide k . Since m and n are coprime, we conclude that g k = (1 , 1) if and only if mn divides k . Thus o ( g ) = mn. Hence ( g ) is a subgroup of order mn . Therefore C m C n = ( g ) = C mn (using Theorem 4.3 to tell us that there is a unique cyclic group of any given order up to isomorphism). square Repeated use of Lemma 6.1 tells us: Corollary 6.2 Let n be a positive integer and write n = p k 1 1 p k 2 2 ...p k r r as a product of prime powers (where the prime numbers p 1 , p 2 , . . . , p r are distinct). Then C n = C p k 1 1 C p k 2 2 C p kr r ....
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 Winter '11
 PhanThuongCang

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