Let $A$ be an abelian torsion-free group. If $A$ is isomorphic with the group of rational numbers whose denominators are powers of, say, $2$, then its automorphism group is isomorphic with $\mathbb{Z}\_2\times\mathbb{Z}$.


Is there an abelian torsion-free group which is not locally cyclic with such automorphism group? And what about (abelian torsion-free non-(locally cyclic)) groups with automorphism groups such as $\mathbb{Z}\_2\times\mathbb{Z}^n$?



Abelian torsion-free group with $\mathbb{Z}_2\times\mathbb{Z}$ as automorphism group
Here's one, a subgroup of $\mathbf{Q}^2$.


Below $\mathbf{Z}\_p$ means the $p$-adic ring.


Let $R$ be the matrix $\begin{pmatrix}0 & 1\\ 1 & 6\end{pmatrix}$. Let $a,b$ be the two roots $X^2-6X-1$ in $\mathbf{Q}\_3$, with $a\equiv 1$ (mod $3$). We have $\mathbf{Q}\_3^2=V\_a\oplus V\_b$, where $V\_a=\mathrm{Ker}(R-a)$ and $V\_b=\mathrm{Ker}(R-b)$. Define $$G=\mathbf{Z}[1/3]^2\cap (\mathbf{Z}\_3^2+V\_a).$$So, $\mathbf{Z}^2\subset G$ and $G/\mathbf{Z}^2$ is isomorphic to $\mathbf{Z}[1/3]/\mathbf{Z}$.


The automorphism group $A$ of $G$ is the set of matrices in $\mathrm{GL}\_2(\mathbf{Z}[1/3])$ preserving $G$. Moreover, $A$ preserves $V\_a$, because $V\_a$ is the "asymptotic direction" of $G$ viewed as subset of $\mathbf{Q}\_3^2$ (more precisely, the Hausdorff limit of $\lambda G$ when $\lambda\to 0$ in $\mathbf{Q}\_3^2$ is $V\_a$; alternatively $V\_a$ is also the intersection of $\lambda\bar{G}$ over $\lambda\in\mathbf{Q}\_3^\*$). For $s\in A$, this means that $s$ fixes $V\_a$ for its action on the projective line, and hence it also fixes its Galois conjugate $V\_b$. In $\mathrm{GL}\_2(\mathbf{Q}\_3)$, the set of matrices preserving each of $V\_a$ and $V\_b$ is just a conjugate of the diagonal subgroup (by some matrix in $\mathrm{GL}\_2(\mathbf{Z}\_3)$, and equals the centralizer of $R$. Hence $A$ is contained in the centralizer of $R$ in $\mathrm{GL}\_2(\mathbf{Z}[1/3])$. Also $A$ preserves $V\_b\cap \mathbf{Z}\_3^2$, so each $s\in A$ acts on $V\_b$ with an eigenvalue in $\mathbf{Z}\_3^\*$; by Galois conjugation it's also true on $V\_a$. Hence $A\subset\mathrm{GL}\_2(\mathbf{Z}\_3)\cap\mathrm{GL}\_2(\mathbf{Z}[1/3])=\mathrm{GL}\_2(\mathbf{Z})$.


I think that $A$ is reduced to $\langle R\rangle\times\{\pm I\_2\}$. Anyway (by laziness) let's check almost this, namely that it's isomorphic to $\mathbf{Z}\times C\_2$. Since $A$ contained in the group of $\mathbf{Z}$-points of a 1-dimensional $\mathbf{Q}$-torus (with 2 components), it is abelian and infinite virtually cyclic. The amalgam structure of $\mathrm{GL}\_2(\mathbf{Z})$ implies that the only such subgroups are isomorphic to $\mathbf{Z}$ or $\mathbf{Z}\times C\_2$. (To conclude that $A$ is reduced to $\langle R\rangle\times\{\pm I\_2\}$ it would be enough to check that $R$ is not a proper power in $\mathrm{GL}\_2(\mathbf{Q})$, which is probably a simple exercise of algebraic number theory.)


I guess that other subgroups of $\mathbf{Q}^n$ defined in similar fashion also achieve automorphism groups of the form $\mathbf{Z}^d\times C\_2$.


