On abelian groups by which balanced extensions of a rational group split

On abelian groups by which balanced extensions of a rational group split. II

“…Proof. Considering that the elements of A not in A**(t) are all of types ^ t, we can argue as in [BF,. We conclude that all members of this separative chain are .R-groups.…”

“…Proof. If A** (t) =0, then we are in the situation of [BF,4.5], and the claim in (i) follows. If A**(t) has rank one, then by (3.5) A is a prebalanced extension rank one group by a rank one group, so it is a Butler group.…”

“…Let 0->P-->G^-»-4-»0bea balanced-exact sequence. We want to define a map xp: G -• » R which is the identity on R. Setting G u = (p~1(A u ), we obtain a chain A**(t) = Go < G\ < ... < G u < ... with union G where the subgroups G u are prebalanced in G. Therefore, for each v, we can write G u +\ = G u + J\ + ... + J^ for a finite number of rank one subgroups J{ of G which are evidently all of types ^ t. [BF,2.1] guarantees that G u is t-cobalanced in G u +\. Condition (a) implies that there is a map xpo: Go ->• I?…”

“…If J is a rank one pure subgroup of B, then by the regularity of K = Ker Sa in B the group 6aJ is finite, and so [BF,4.2,4.3] guarantees the balancedness of the top row. To complete the proof, we must show that the top row does not split.…”

“…Suppose A is as stated and is an .R-group. (a) and (b) follow from (3.1) and (3.4), respectively, while (c) is a consequence of (1.4), (1.5), and [BF,7.1].…”