Inertial endomorphisms of an abelian group. (Q5964958)

Let \(A\) be an abelian group. An endomorphism \(\varphi\colon A\to A\) is \textit{(right) inertial} if \((X+\varphi(X))/X\) is finite for every subgroup \(X\) of \(A\). The inertial endomorphisms of \(A\) form a ring, denoted by \(\text{IE}(A)\), which contains the ideal \(F(A)\) of all finitary (i.e., with finite image) endomorphisms of \(A\). A consequence of the main theorem of this paper, which characterizes the inertial endomorphisms of \(A\), is that \(\text{IE}(A)/F(A)\) is a commutative ring. Let now \(\text{IAut}(A)\) be the group generated by all inertial automorphisms of \(A\), which contains the locally finite group \(\text{FAut}(A)\) of all finitary automorphisms of \(A\). The derived subgroup of \(\text{IAut}(A)\) is proved to be contained in \(\text{FAut}(A)\). Consequently, \(\text{IAut}(A)\) is locally central-by-finite and \(\text{IAut}(A)/\text{FAut}(A)\) is a commutative group. An endomorphism \(\varphi\colon A\to A\) is \textit{left inertial} if \((X+\varphi(X))/\varphi(X)\) is finite for every subgroup \(X\) of \(A\). Clearly, if \(\varphi\) is a right inertial automorphism of \(A\), then \(\varphi^{-1}\) is left inertial. When \(A\) has finite torsion-free rank, it is proven that a left inertial endomorphism \(\varphi\) of \(A\) is necessarily right inertial; in case \(\varphi\) is an automorphism of \(A\), then \(\varphi\) is right inertial if and only if \(\varphi\) is left inertial. So, if \(A\) has finite torsion-free rank, the elements of the group \(\text{IAut}(A)\) are precisely all inertial automorphisms of \(A\). Moreover, if \(A\) has infinite torsion-free rank, then an endomorphism \(\varphi\) of \(A\) is both right and left inertial precisely when \(\varphi\) acts as \(id\) or \(-id\) on a finite-index subgroup of \(A\).