On pⁿBext projective abelian p-groups.pdf 1. Introduction and Background Everywhere in the text of this brief paper our groups are p-primary abelian, where p is a fixed prime for the duration of the article. The undefined explicitly below notions and notations are in agreement with [5]. For instance, a group G is called balanced projective if the equality Bext(G, X) = {0} holds for all groups X. In order to generalize this, imitating [3], for any integer n > 0, we say that the short exact sequence 0 ^ X ^ Y ^ G ^ 0 is n-balanced exact if it represents an element of pnBext(G, X). Thus we will say that a group G is n-balanced projective provided every such n-balanced exact sequence splits. Evidently, these two notions coincide when n = 0. It is worthwhile noticing that certain non-trivial properties of these groups are given in [3] (see also [4]). These ideas lead us to the next new concept: Definition 1.1. Let n > 0 . A group G is said to be pnBext-projective if (VX), pn - Bext(G, X) = {0}. The aim of this note is to prove that each n-balanced projective group is pn yields Bext-projective but the converse fails. We close the work with a specific question arisen from unexpected difficulties in the proof of the central statement. 2. Main Result and Problem Theorem 2.1. Suppose that G is a group and n 2 does there exist a pnBext-projective group that is not n-totally projective? Resuming, we have restricted our attention only on n = 1, though essentially the same argument works for larger values of n (see cf. [1] and [2] too). In fact, last argument stated above asserts that any element of pn Bext( A, X )will be n-balanced exact, so that every group which is projective with respect to the collection of n-balanced exact sequences will also be projective with respect to the functor pnBext. The second assertion then implies that there are n-balanced exact sequences that are not elements of pnBext(A, X). Besides, notice that the totally projective (i.e., the balanced projective) groups are exactly p0Bext-projective groups, and there are an abundance of them. Nevertheless, it is actually not at all clear whether there are enoughpnBext-projectives whenever n > 0. So, the following homological question is of some interest: Problem. Is it true that the collection of n-balanced exact sequences form the largest subfunctor of pnBext which does have this important homological property?

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