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.The proof of this theorem can be found in any of the references atthe end of this chapter.Theorem 11.5 (Fundamental Theorem of Finitely Generated AbelianGroups) Every finitely generated abelian group G is isomorphic to a directproduct of cyclic groups of the formZ α ×α× · · · ××p 1Zp 2ZpαnZ × · · · × Z,12nwhere the pi’s are primes (not necessarily distinct).11.2Solvable GroupsA subnormal series of a group G is a finite sequence of subgroupsG = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e},where Hi is a normal subgroup of Hi+1.If each subgroup Hi is normal inG, then the series is called a normal series.The length of a subnormalor normal series is the number of proper inclusions.Example 4.Any series of subgroups of an abelian group is a normal series.Consider the following series of groups:Z ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0},Z24 ⊃ h2i ⊃ h6i ⊃ h12i ⊃ {0}.Example 5.A subnormal series need not be a normal series.Consider thefollowing subnormal series of the group D4:D4 ⊃ {(1), (12)(34), (13)(24), (14)(23)} ⊃ {(1), (12)(34)} ⊃ {(1)}.The subgroup {(1), (12)(34)} is not normal in D4; consequently, this seriesis not a normal series.A subnormal (normal) series {Kj} is a refinement of a subnormal(normal) series {Hi} if {Hi} ⊂ {Kj}.That is, each Hi is one of the Kj.196CHAPTER 11THE STRUCTURE OF GROUPSExample 6.The seriesZ ⊃ 3Z ⊃ 9Z ⊃ 45Z ⊃ 90Z ⊃ 180Z ⊃ {0}is a refinement of the seriesZ ⊃ 9Z ⊃ 45Z ⊃ 180Z ⊃ {0}.The correct way to study a subnormal or normal series of subgroups,{Hi} of G, is actually to study the factor groups Hi+1/Hi.We say that twosubnormal (normal) series {Hi} and {Kj} of a group G are isomorphic ifthere is a one-to-one correspondence between the collections of factor groups{Hi+1/Hi} and {Kj+1/Kj}.Example 7.The two normal seriesZ60 ⊃ h3i ⊃ h15i ⊃ {0}Z60 ⊃ h4i ⊃ h20i ⊃ {0}of the group Z60 are isomorphic sinceZ60/h3i ∼= h20i/{0} ∼= Z3h3i/h15i ∼= h4i/h20i ∼= Z5h15i/{0} ∼= Z60/h4i ∼= Z4.A subnormal series {Hi} of a group G is a composition series if all thefactor groups are simple; that is, if none of the factor groups of the seriescontains a normal subgroup.A normal series {Hi} of G is a principalseries if all the factor groups are simple.Example 8.The group Z60 has a composition seriesZ60 ⊃ h3i ⊃ h15i ⊃ h30i ⊃ {0}with factor groupsZ60/h3i∼=Z3h3i/h15i ∼=Z5h15i/h30i ∼=Z2h30i/{0} ∼=Z2.11.2SOLVABLE GROUPS197Since Z60 is an abelian group, this series is automatically a principal series.Notice that a composition series need not be unique.The seriesZ60 ⊃ h2i ⊃ h4i ⊃ h20i ⊃ {0}is also a composition series.Example 9.For n ≥ 5, the seriesSn ⊃ An ⊃ {(1)}is a composition series for S∼n since Sn/An = Z2 and An is simple.Example 10.Not every group has a composition series or a principal series.Suppose that{0} = H0 ⊂ H1 ⊂ · · · ⊂ Hn−1 ⊂ Hn = Zis a subnormal series for the integers under addition.Then H1 must be ofthe form n∼Z for some n ∈ N.In this case H1/H0 = nZ is an infinite cyclicgroup with many nontrivial proper normal subgroups.Although composition series need not be unique as in the case of Z60, itturns out that any two composition series are related.The factor groups ofthe two composition series for Z60 are Z2, Z2, Z3, and Z5; that is, the twocomposition series are isomorphic.The Jordan-Hölder Theorem says thatthis is always the case.Theorem 11.6 (Jordan-Hölder) Any two composition series of G areisomorphic.Proof.We shall employ mathematical induction on the length of the com-position series.If the length of a composition series is 1, then G must be asimple group.In this case any two composition series are isomorphic.Suppose now that the theorem is true for all groups having a compositionseries of length k, where 1 ≤ k < n.LetG = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}G = Km ⊃ Km−1 ⊃ · · · ⊃ K1 ⊃ K0 = {e}be two composition series for G.We can form two new subnormal series forG since Hi ∩ Km−1 is normal in Hi+1 ∩ Km−1 and Kj ∩ Hn−1 is normal inKj+1 ∩ Hn−1:G = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e}G = Km ⊃ Km−1 ⊃ Km−1 ∩ Hn−1 ⊃ · · · ⊃ K0 ∩ Hn−1 = {e}.198CHAPTER 11THE STRUCTURE OF GROUPSSince Hi∩Km−1 is normal in Hi+1∩Km−1, the Second Isomorphism Theoremimplies that(Hi+1 ∩ Km−1)/(Hi ∩ Km−1) = (Hi+1 ∩ Km−1)/(Hi ∩ (Hi+1 ∩ Km−1))∼=Hi(Hi+1 ∩ Km−1)/Hi,where Hi is normal in Hi(Hi+1 ∩ Km−1).Since {Hi} is a composition se-ries, Hi+1/Hi must be simple; consequently, Hi(Hi+1 ∩ Km−1)/Hi is eitherHi+1/Hi or Hi/Hi.That is, Hi(Hi+1 ∩ Km−1) must be either Hi or Hi+1.Removing any nonproper inclusions from the seriesHn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e},we have a composition series for Hn−1.Our induction hypothesis says thatthis series must be equivalent to the composition seriesHn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}.Hence, the composition seriesG = Hn ⊃ Hn−1 ⊃ · · · ⊃ H1 ⊃ H0 = {e}andG = Hn ⊃ Hn−1 ⊃ Hn−1 ∩ Km−1 ⊃ · · · ⊃ H0 ∩ Km−1 = {e}are equivalent.If Hn−1 = Km−1, then the composition series {Hi} and {Kj}are equivalent and we are done; otherwise, Hn−1Km−1 is a normal subgroupof G properly containing Hn−1.In this case Hn−1Km−1 = G and we canapply the Second Isomorphism Theorem once again; that is,Km−1/(Km−1 ∩ Hn−1) ∼= (Hn−1Km−1)/Hn−1 = G/Hn−1
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