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216 ULRICH ALBRECHTPROPOSITION 3.2. Let A be a self-small abelian group which is flatas an E(A)-module. The following conditions are equivalent for anabelian group G with S A(G) = G:(a) G is A-solvable.n(b) IfO-+U^>P-!->G->Oisan exact sequence such that P isA-projective, then S A(U) = U.(c) Every epimorphism ΦQ: PQ-^> G where PQ is A-projective extendsto a long exact sequenceΦn+l p Φn Φ\ p Φθ ^ rχ• • Γ n• * * * • ΓQ —• LΓ —• Uin which P nis A-projective for all n < ω.(d) There exists an exact sequence -^ P nh Λ P oH G -• 0such that, for all n <ω, P nis A-projective and the induced sequenceis A-balanced.aProof, (a) => (b): The exact sequence 0->C/AP^G-^0 inducesa projective resolution0 -> H A(U) *H α) H A(P) HA X β) M -> 0of the right £'(y4)-module M = imH A(β). By Lemma 3.1, there isa map 0(/?): T A(M) —• G which fits into the following commutativediagram whose rows are exact:0-+T A{M) - T AH A{G)IT A(M)Θ( Λ ] G-+0Thus, θ(β) is an isomorphism, and S A(U) = U by Lemma 3.1.(b) => (c): The map φo induces an exact sequencewhere S A(U) = C/ because of (b). Lemma 3.1 yields that U is A-solvable. The long exact sequence is now constructed inductively.(c) => (d): There exists an ^-balanced exact sequence0- u^Po^G-*O(l)

LOCALLY ^-PROJECTIVE ABELIAN GROUPS AND GENERALIZATIONS 217where PQ is ^4-projective, and U c P osatisfies S A(U) = U because of(c). Since U is ^4-solvable by Lemma 3.1, and we have already verifiedthe implication (a) => (c), we obtain an ^4-balanced exact sequencewith Pi ^4-projective. Because of the validity of the implication (a) =>*(b), S A(V) = V. Consequently, an inductive argument completes theproof.(d) =» (a): The sequence in (d) induces an exact sequence0 - T AH A(U)Θ^U^ T AH A{G) Θ Λ G -> 0by Lemma 3.1 where U = imφ\. Because of S A(U) = U 9the map θjjis onto; and ΘQ is an isomorphism.EXAMPLE 3.3. Let A be a countable abelian group of infinite rankwith E(A) = Z. Every free subgroup F of A which has infinite rankcontains a subgroup F xsuch that F/F x= T A(Q) = 0 ωQ. Thus,F/F\ is a direct summand of A/F\ 9and there exists a non-zero propersubgroup U of A with A/U = 0 ωQ.We now show that S A(U) = 0. If this is not the case, then H A(U) Φ0; and H A(A)/H A(U) is a bounded abelian group. However, since thelatter is isomorphic to a subgroup of the torsion-free group H A(® ωQ),this is only possible if H A{A) = H A(U). Because this contradicts thecondition A Φ U 9we obtain S A(U) = 0. On the other hand, every freeresolution 0 —• 0 ωZ -• 0 ωZ -• Q -• 0 yields an exact sequenceFinally, we construct an 4-balanced exact sequence 0 -> V —•φ 74 -• 0 ωQ -^ 0 such that S A{V) φ V: For this, we observeHA(Q) — Θ2 N o Q Hence, there exists an 4-balanced exact sequence0 -> F ^ 0 2κ o4 -^ 0 ωQ -• 0. By Proposition 3.2, ^(F) ^ F.In the next part of this section, we introduce the concept of an A-projective dimension for 4-solvable groups. In view of Proposition3.2, we assume, that A is self-small and flat as an 2s(4)-module, andconsider two 4-balanced 4-projective resolutions 0 —• I// —• PiG —> 0(/ = 1,2) of an ^4-solvable group G. They induce exact sequences0 -• H A(Ui) -+ H Λ(Pi) -> H A{G) -+ 0 of right E{A)-moάxx\es for/ = 1,2. By ShanueΓs Lemma [R, Theorem 3.62], H A(P {) ®H A(U 2) =HA(PI)®H A(U\). Since both, the C//'s and the P/'s, are ^-solvable, weω

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