In this note we provide a different proof of Hill's criteria of freeness for abelian groups. Our proof hinges on the construction of suitable $G(\aleph_0)$-families of subgroups of the links in Hill's theorem and, ultimately, on the construction of such a family of pure subgroups of the group itself

Macías-Díaz, Jorge Eduardo

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Revista Colombiana de MatemáticasVolumen 44(2010)1, páginas 59-64An Alternative Proof of Hill’s Criterionof Freeness for Abelian GroupsUna prueba alternativa del criterio de Hill para grupos abelianoslibresJorge Eduardo Maćıas-Dı́azUniversidad Autónoma de Aguascalientes, Aguascalientes, MéxicoAbstract. In this note we provide a different proof of Hill’s criteria of freenessfor abelian groups. Our proof hinges on the construction of suitable G(ℵ0)-families of subgroups of the links in Hill’s theorem and, ultimately, on theconstruction of such a family of pure subgroups of the group itself.Key words and phrases. Abelian group, Freeness, Hill’s criterion, G(ℵ0)-family,Purity.2000 Mathematics Subject Classification. 20K20, 03E75, 20K25.Resumen. En este trabajo se proporciona una nueva demostración del criteriode Hill para grupos abelianos libres. La demostración se basa en la construc-ción de una G(ℵ0)-familia de subgrupos en los eslabones del teorema de Hilly, prioritariamente, en la construcción de una familia tal de subgrupos puros.Palabras y frases clave. Grupo abeliano, libertad, criterio de Hill, G(ℵ0)-familia,pureza.1. IntroductionIn 1934, Lev Pontryagin proved that a countable, torsion-free abelian groupis free if and only if every finite rank, pure subgroup is free [3]. Equivalently,every properly ascending chain of subgroups of the same finite rank is finite.From the proof of this criterion, it follows that a torsion-free abelian group Gis free if there exists an ascending chain0 = G0 < G1 < · · · < Gn < · · · , (n < ω), (1)consisting of pure subgroups of G whose union is equal to G, such that everyGn is free and countable. Here, a subgroup H of the abelian group G is pure5960 JORGE EDUARDO MAĆıAS-Dı́AZif solubility in G of every equation of the form nx = h ∈ H , with n ∈ Z,implies its solubility in H . Also, we say that G is torsion-free if n = 0 or g = 0,whenever n ∈ Z and g ∈ G satisfy ng = 0.Later, in 1970, Hill established that, in order for an abelian group G to befree, it is sufficient to prove that it is the union of a countable ascending chain(1) consisting of free, pure subgroups [1]. In other words, he proved the followingtheorem, establishing thus that the countability condition on the cardinality ofthe links of the chain was superfluous.Theorem 1 (Hill’s criterion of freeness). A torsion-free abelian group G is freeif there exists a countable ascending chain0 = G0 < G1 < · · · < Gn < · · · , (n < ω) (2)of subgroups of G, such that:a) every Gn is free,b) every Gn is a pure subgroup of G, andc) G =⋃n<ω Gn.In this note, we give a proof of Hill’s criterion different from the one providedin [1]. Our proof hinges on the construction of suitable classes of subgroups ofthe groups Gn and, ultimately, on the construction of such a family consistingof pure subgroups of G. Section 3 of this work contains the proof of Theorem1, while Section 2 presents some preliminary results.2. Preparatory LemmasThe following is a general result which will be used in the proof of Theorem 1.We refer to [2] for definitions of the set-theoretical concepts.Lemma 1. An abelian group G is free if there exists a continuous, well-ordered,ascending chain0 = A0 < A1 < · · · < Aγ < Aγ+1 < · · · , (γ < τ) (3)of subgroups of G, such that:a) every factor group Aγ+1/Aγ is free, andb) G =⋃γ<τ Aγ .Proof. The conclusion follows from the fact that G is isomorphic to the directsum of the factor groups Aγ+1/Aγ , for γ < τ . XVolumen 44, Número 1, Año 2010ANOTHER PROOF OF HILL’S CRITERION 61Recall that a G(ℵ0)-family of an abelian group G is a collection B of sub-groups of G, which satisfies the following properties:i) 0 and G belong to B,ii) B is closed under unions of ascending chains, andiii) for every A0 ∈ B and every countable set H ⊆ G, there exists A ∈ B whichcontains both A0 and H , such that A/A0 is countable.Clearly, every abelian group has a G(ℵ0)-family, namely, the collection ofall its subgroups.For the rest of this section, we will assume the hypotheses of Theorem 1.Under these circumstances, we fix a basis Xn of Gn for every n < ω, and letBn be the family of all subgroups of Gn generated by subsets of Xn. Clearly,every member of Gn is a direct summand of Gn and, thus, a pure subgroup ofG.Lemma 2. The collection B′n = {A ∈ Bn | A+Gi is pure in G, for every i <ω} is a G(ℵ0)-family of pure subgroups of Gn, for every n < ω.Proof. All we need to check is that the countability condition is satisfied, sincethe other conditions of a G(ℵ0)-family are obvious. So, let A0 ∈ B′n, and let H0be a countable subset of Gn. Moreover, let m < ω, and assume that we havealready constructed a chainA0 < A1 < · · · < Am (4)of groups in Bn, such that:a) H0 is contained in A1,b) for every j < m, the group Aj+1/Aj is countable, andc) for every j < m and every i < ω, (Aj+1 + Gi)/(A0 + Gi) contains thepurification of (Aj +Gi)/(A0 +Gi) in G/(A0 +Gi).To find the next member of (4), for every i < ω, let Vi ⊆ Gn be a completeset of representatives of the purification of (Am+Gi)/(A0+Gi) in G/(A0+Gi).The sets Vi are clearly countable, so that Hm+1 = H0 ∪⋃i<ω Vi is likewisecountable. Therefore, there exists Am+1 ∈ Bn containing both Am and Hm+1,such that Am+1/Am is countable. Inductively, we construct a chainA0 < A1 < · · · < Am < · · · , (m < ω) (5)of groups in Bn, satisfying properties a), b) and c) above, for every m < ω.Revista Colombiana de Matemáticas62 JORGE EDUARDO MAĆıAS-Dı́AZEvidently, the union A of the links of (5) is a member of Bn, A/A0 iscountable, and our construction guarantees that (A + Gi)/(A0 + Gi) is purein G/(A0 + Gi). Thus, A + Gi is pure in G and, consequently, A belongs toB′n. XLemma 3. The collection B = {A < G | A ∩ Gn ∈ B′n, for every n < ω} is aG(ℵ0)-family of pure subgroups of G.Proof. Again, only the countability condition merits attention; so, let A0 ∈ B,and let H ⊆ G be countable. For every k < ω, let A0k = A0 ∩Gk. Moreover, letn < ω, and assume that we have already constructed a finite ascending chainA0 < A1 < · · · < An (6)of subgroups of G, such that all factor groups Am/A0 are countable, for everym ≤ n. Furthermore, suppose that each link Am in (6) may be expressed asthe union of a countable ascending chain0 = Am0 < Am1 < · · · < Amk < · · · , (k < ω) (7)of subgroups of G, such that:a) Amk ∈ B′k, for every k < ω and every m ≤ n,b) Amk is countable over A0 ∩Gk, for every k < ω and every m ≤ n, andc) Amk < Am ∩Gk < Am+1k , for every k < ω and m+ 1 ≤ n.For every k < ω, the group (An ∩ Gk)/(A0 ∩ Gk) is countable, so we mayfix a countable set of representatives Yk of An∩Gk modulo A0∩Gk. Moreover,there exists Bk ∈ B′k containing both A0∩Gk and Yk, such that Bk is countableover A0 ∩ Gk. Thus, any set of representatives Hk of Bk modulo A0 ∩ Gk iscountable.In order to construct the next link in (6), assume that the groups in theascending chain 0 = An+10 < An+11 < · · · < An+1k have been built as needed,for some k < ω, and let Zk ⊆ Gk be a set of representatives of Am+1k moduloA0 ∩ Gk. Then, there exists An+1k+1 ∈ B′k+1 which contains A0 ∩ Gk+1 and thecountable set Zk ∪Hk+1 ∪ (H ∩Gk+1), such that An+1k+1 is countable over A0 ∩Gk+1.Clearly, the group A =⋃n<ω An contains both A0 and H , and is countableover A0. Moreover, our construction guarantees that A ∩ Gk ∈ Bk, for everyk < ω. We conclude that A ∈ B. XBefore we prove our next result, it is important to notice that A + Gnis a pure subgroup of G, for every A ∈ B and every n < ω. Indeed, thatVolumen 44, Número 1, Año 2010ANOTHER PROOF OF HILL’S CRITERION 63(A + Gn) ∩ Gn+1 is pure in G follows from the fact that A ∩ Gn+1 ∈ B′n+1.Next, assume that (A + Gn) ∩ Gk is pure in G, for some k > n. It is easy tocheck that(A+Gk) ∩Gk+1(A+Gn) ∩Gk+1∼=Gk(A+Gn) ∩Gk, (8)whence it follows that (A + Gn) ∩ Gk+1 is pure in G. The claim is readilyestablished after noticing that A+Gn =⋃k<ω(A+Gn) ∩Gk.Lemma 4. For every A ∈ B, finite rank, pure subgroups of G/A are free.Proof. Let A ∈ B, and let D be a pure subgroup of G containing A, such thatD/A is of finite rank. If S = {d1, . . . , dn} is a complete set of representatives ofa maximal independent system of D modulo A, then there exists k < ω suchthat S ⊆ Gk. Then A + (D ∩ Gk) = D ∩ (A + Gk) is a pure subgroup of Gcontaining S, which lies between A and D. Therefore, D = A+ (D ∩Gk). Thefact that A ∩ Gk ∈ B′k implies that A ∩ Gk is a summand of Gk. Therefore,there exists a finite rank, free group B, such that D ∩ Gk = (A ∩ Gk) ⊕ B.Notice thatD = A+ (D ∩Gk) = A+((A ∩Gk)⊕B)= A⊕B, (9)which implies that D/A is free. X3. Proof of the Main ResultProof of Theorem 1. Let α be any nonzero ordinal, and let0 = A0 < A1 < · · · < Aγ < Aγ+1 · · · , (γ < α) (10)be an ascending chain of subgroups in B, such that all factor groups Aγ+1/Aγare free. If α is a limit ordinal, then we let Aα =⋃γ<αAγ . Otherwise, thereexists an ordinal β such that α = β+1. In this case, if there exists x ∈ G \Aβ,we let Aβ+1 ∈ B contain both x and Aβ , such that Aβ+1/Aβ be countable.Lemma 4 implies now that finite rank, pure subgroups of Aβ+1/Aβ are free.Consequently, Aβ+1/Aβ is free by Pontryagin’s criterion.Using transfinite induction, we construct a continuous, well-ordered, ascend-ing chain (3) of subgroups of G satisfying properties (a) and (b) of Lemma 1.We conclude that G is free. XReferences[1] P. Hill, On the Freeness of Abelian Groups: A Generalization of Pontryagin’sTheorem, Bullet. Amer. Math. Soc. 76 (1970), no. 5, 1118–1120.Revista Colombiana de Matemáticas64 JORGE EDUARDO MAĆıAS-Dı́AZ[2] T. Jech, Set Theory, Springer Monographs in Mathematics, Springer-Verlag,Berlin, Germany, 3th ed., 2003.[3] L. Pontryagin, The Theory of Topological Commutative Groups, Annals ofMath. 35 (1934), no. 2, 361–388.(Recibido en julio de 2008. Aceptado en abril de 2010)Departamento de Matemáticas y F́ısicaUniversidad Autónoma de AguascalientesAvenida Universidad 940Ciudad UniversitariaAguascalientes, Ags. 20131Méxicoe-mail: jemacias@correo.uaa.mxVolumen 44, Número 1, Año 2010

An alternative proof of hill's criterion of freeness for abelian groups

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