Universally fully and Krylov transitive torsion-free abelian groups (Q2145916)

A group \(G\) is called \textit{Krylov transitive} if, for any elements \( x,y\in G\) with \(\chi _{G}(x)=\chi _{G}(y)\) (\(\chi _{G}(x)\) denotes the characteristic), there exists an endomorphism of \(G\) mapping \(x\) to \(y\). A group \(G\) is \textit{universally fully transitive} if, for every group \(M\) and every pair of elements \(x\in G\) and \(y\in M\), if \(\chi _{G}(x)\leq \chi _{M}(y)\) then there is a homomorphism \(\phi :G\rightarrow M\) such that \(\phi (x)=y\). If this holds whenever \(\chi _{G}(x)=\chi _{M}(y)\), the group G is said to be universally Krylov transitive. Such groups were characterized in the torsion case by the same authors in a previous paper [J. Algebra 566, 187--204 (2021; Zbl 1473.20056)]. In the torsion-free case, these turn out to be the same. We browse from the results. Theorem 2.1. Suppose \(G\) is a group and \(G=R\oplus D\), where \(R\) is reduced and \(D\) is divisible. Then the following are equivalent: \begin{itemize} \item[(a)] \(G\) is universally fully transitive. \item[(b)] \(G\) is universally Krylov transitive. \item[(c)] Every pure rank-1 subgroup of \(G\) is a summand. \item[(d)] Every pure finite-rank subgroup of \(G\) is a summand. \item[(g)] \(R\) is a homogeneous separable group (and three other conditions). \end{itemize} Corollary 2.4. If \(G\) is a pure subgroup of a direct product of copies of \(\mathbb{Z}\), then \(G\) is universally fully transitive. In particular, the Baer-Specker group is universally fully transitive. Corollary 2.11. The class of \(\sigma \)-homogeneous universally fully transitive groups is the smallest class containing \(\mathbb{Q}_{\rho }\) that is closed with respect to direct products and pure subgroups. Theorem 2.13. Suppose \(\sigma =\overline{\rho }\) is an idempotent type. If \(G\) is a \(\mathbb{Q}_{\rho }\)-module and \(G=R\oplus D\), where \(D\) is divisible and \(R\) is reduced, then the following are equivalent: \begin{itemize} \item[(a)] \(G\) is universally fully transitive and if \(R=0\), then it has a pure submodule isomorphic to \(\mathbb{Q}_{\rho }\). \item[(b)] \(G\) is universally Krylov transitive and if \(R=0\), then it has a pure submodule isomorphic to \(\mathbb{Q}_{\rho }\). \item[(c)] \(R\) is isomorphic to a pure submodule of some direct product \( \prod\limits_{i\in I}\mathbb{Q}_{\rho }\). \item[(d)] Under the natural map \(\phi _{R}:R\rightarrow R^{\ast \ast }\), \(R\) maps to a pure submodule of \(R^{\ast \ast }\). \end{itemize} Section 3 deals with a relativization to a subgroup \(H\) of a group \(G\): \(H\)-fully and \(H\)-Krylov transitive groups, which are (initially) defined as follows: Given \(H\leq G\), we say that the group \(G\) is \(H\)\textsl{-fully transitive} if, for every pair of elements \(x\in H\) and \(y\in G\), if \(\chi _{H}(x)\leq \chi _{G}(y)\) then there is a homomorphism \(f:H\rightarrow G\) such that \(f(x)=y\). If this holds when \(\chi _{H}(x)=\chi _{G}(y)\), then the group \(G\) is said to be \(H\)\textsl{-Krylov transitive}. However, in this section \(H\) is not supposed to be a subgroup of \(G\), but an arbitrary fixed group. Some of the results are: Proposition 3.5. Let \(G=\bigoplus\limits_{i\in I}G_{i}\) be a torsion-free group, where \(\pi _{i}:G\rightarrow G_{i}\) are the corresponding projections, and let \(H\leq G\) be an invariant subgroup, comparatively on the system \(\{\pi _{i}\}_{i\in I}\) , i.e., \( H=\bigoplus\limits_{i\in I}H_{i}\) , where \(H_{i}=\pi _{i}(H)\). Then \(G\) is \(H \)-fully transitive if, and only if, the system \(\{G_{i}\}_{i\in I}\) satisfies the monotonic condition with respect to the characteristics of the subgroups \(\{H_{i}\}_{i\in I}\) and, for all indices \(i,j\in I\), the groups \( G_{i}\), \(G_{j}\) form a completely transitive pair of the corresponding subgroups \(H_{i}\), \(H_{j}\). Corollary 3.7. Suppose \(G=D(G)\oplus A\), where \(D(G)\) is the divisible part of \(G\) and suppose \(H=F\oplus R\) is a subgroup of \(G\), where \( F\leq D(G)\), \(R\leq A\). If \(G\) is \(H\)-fully transitive, then \(A\) is both \(R\) -fully transitive and \(F\)-fully transitive. The converse implication fails in general. Proposition 3.11. A vector group is universally fully transitive if and only if all of the rank 1 groups forming the direct product are of the same idempotent type. The paper ends with a section ``Concluding discussion and open problems''.