Abstract Algebra Theory and Applications - Computer Science ...

192 CHAPTER 11 THE STRUCTURE OF GROUPSClearly, the set K is closed under the group operation. Since gi 0 = 1,the identity is in K. It remains to show that the inverse of an elementg = g k 11 · · · gkn i nin K must also be in K. However,g −1 = (g k 11 · · · gkn i n) −1 = (g −kn1 · · · g −k 1i n).Now let us restrict our attention to finite abelian groups. We can expressany finite abelian group as a finite direct product of cyclic groups. Morespecifically, letting p be prime, we define a group G to be a p-group if everyelement in G has as its order a power of p. For example, both Z 2 × Z 2 andZ 4 are 2-groups, whereas Z 27 is a 3-group. We shall prove that every finiteabelian group is isomorphic to a direct product of cyclic p-groups. Before westate the main theorem concerning finite abelian groups, we shall considera special case.Theorem 11.2 Every finite abelian group G is the direct product of p-groups.Proof. If |G| = 1, then the theorem is trivial. Suppose that the order ofG is greater than 1, say|G| = p α 11 · · · pαn n ,where p 1 , . . . , p n are all prime, and define G i to be the set of elements in G oforder p k i for some integer k. Since G is an abelian group, we are guaranteedthat G i is a subgroup of G for i = 1, . . . , n. We must show thatG = G 1 × · · · × G n .□That is, we must be able to write every g ∈ G as a unique product g p1 · · · g pnwhere g pi is of the order of some power of p i . Since the order of g dividesthe order of G, we know that|g| = p β 11 pβ 22 · · · pβn nfor integers β 1 , . . . , β n . Letting a i = |g|/p β ii , the a i’s are relatively prime;hence, there exist integers b 1 , . . . , b n such that a 1 b 1 + · · · + a n b n = 1. Consequently,g = g a 1b 1 +···+a nb n= g a 1b1· · · g anbn .Sinceg (a ib i )p β ii = g b i|g| = e,