By Perrine C. D.

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Proof: For the sake of the argument let E = End(G). Since Θ and Φ are natural transformations ΘH⊕K = ΘH ⊕ ΘK for groups H and K, and ΦM ⊕N = ΦM ⊕ ΦN for right End(G)-modules M and N . Moreover, since G is a group, for each integer n > 0 there is a group homomorphism ΘG(n) = ⊕n ΘG : TG HG (G(n) ) −→ G(n) . Thus given H ⊕ K ∼ = G(I) we can prove that ΘH is an isomorphism if we can prove that ΘG is an isomorphism. Similarly to show that ΦM is an isomorphism it suffices to show that ΦE is an isomorphism.

Then ΦE (1) = f which implies that ΦE is an isomorphism. 4. ARNOLD-LADY-MURLEY THEOREM 31 The reader proved in Chapter 1 that ΘTG (E) ◦ TG (ΦE ) = 1TG (E) . Then ΘTG (E) is an isomorphism since ΦE and TG (ΦE ) are isomorphisms. Inasmuch as G ∼ = TG (E), ΘG is an isomorphism. Given our reductions, the proof is complete. Let P(G) = {groups H H ⊕ H ∼ = G(c) for some group H and some cardinal c}. Let P(End(G)) be the category of projective right End(G)-modules. 2 Let G = ⊕p Zp where p ranges over the primes in Z.

Dedekind domains are classical maximal orders. If E is a classical maximal order and if U is an E-lattice then Matn (E) and EndE (U ) are classical maximal orders. 1]. 1 Suppose that the rtffr ring E is a classical maximal order. 1. If I is a right ideal of finite index in E then O(I) = {q ∈ QE qI ⊂ I} is a classical maximal order. 16 CHAPTER 1. NOTATION AND PRELIMINARY RESULTS 2. E is an hereditary Noetherian ring. That is, each one-sided ideal in E is a finitely generated projective E-module.