On kernels of purifiability in arbitrary Abelian groups (Q5937623)

The paper deals with arbitrary Abelian groups and basically asks which subgroups are \(p\)-purifiable, i.e., contained in a minimal \(p\)-pure subgroup. The question is not easily answered and has spawned companion problems. A kernel of \(p\)-purity is a subgroup whose neat hulls are \(p\)-pure and a kernel of \(p\)-purifiability is a subgroup whose neat hulls are \(p\)-purifiable. There are groups with subgroups that are kernels of \(p\)-purifiability but not kernels of \(p\)-purity while the kernels of \(p\)-purity obviously are kernels of \(p\)-purifiability. The author obtains the following major results.NEWLINENEWLINENEWLINETheorem~2.14. A subgroup \(A\) of an arbitrary Abelian group \(G\) is a kernel of \(p\)-purity in \(G\) if and only if either (1) \(A[p]\) is \(p\)-dense in \(G[p]\), or (2) for some non-negative integer \(k\), \(G[p]+A=p^kG[p]+A\neq p^{k+1}G[p]+A\) and \(p^n(G/A)[p]=(p^nG[p]+A)/A\) for all \(n>k\).NEWLINENEWLINENEWLINECorollary~2.15. Suppose that there is a non-negative integer such that either (1) \(p^n G[p]+A=p^{n+1}G[p]+A\) for all \(n\geq m\), or (2) \(p^n(G/A)[p]=(p^nG[p]+A)/A\) for all \(n\geq m\). Then \(A\) is \(p\)-purifiable in \(G\).NEWLINENEWLINENEWLINEA subgroup \(A\) of \(G\) is \(p\)-vertical if \((A+p^{n+1}G)\cap p^n G[p]=(A\cap p^nG)[p]+p^{n+1}G[p]\) for all \(n\geq 0\).NEWLINENEWLINENEWLINETheorem~3.6. Let \(G\) be an arbitrary Abelian group and \(A\) a \(p\)-vertical subgroup \(G\). Then \(A\) is a kernel of \(p\)-purifiability of \(G\) if and only if one of the following three conditions holds. (1) \(A\cap p^mG\) is \(p\)-dense in \(p^mG\) for some \(m\geq 0\); (2) \(p^n(G/A)[p]=(p^nG[p]+A)/A\) for some \(m\geq 0\) and all \(n\geq m\); (3) there exist \(m\geq 0\), and subgroups \(H\), \(K\) of \(G\) such that NEWLINE\[NEWLINE\frac{p^mG}{A\cap p^mG}=\frac{H}{A\cap p^mG}\oplus\frac{K}{A\cap p^m G},NEWLINE\]NEWLINE where \(\frac{H}{A\cap p^mG}\) is a divisible subgroup of \(\left(\frac{G}{A\cap p^mG}\right)_p\) of finite rank and NEWLINE\[NEWLINE\frac{K}{A\cap p^mG}[p]=\frac{p^mG[p]+\left(A\cap p^mG\right)}{A\cap p^mG}\text{ such that }p^\omega\left(\frac{K}{A\cap p^m G}\right)[p]=0.NEWLINE\]