# Abelian torsion-free group with $\mathbb{Z}_2\times\mathbb{Z}$ as automorphism group

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 $$\mathbb{Z}_2\times\mathbb{Z}^n$$?

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 $$\mathbf{Z}^d\times C_2$$.

• (Corrected comment) If $A \in \text{GL}_2(\mathbb{Z})$ has no eigenvalue in $\{\pm 1\}$ and is such that $\mathbb{Z}^2$ is cyclic $\mathbb{Z}[A]$-module, then $\langle A \rangle \times \{\pm I_2\}$ is the centralizer of $A$, with two exceptions: if $\text{trace}(A) = \pm 3$, in which case it has index $2$ in the centralizer of $A$. – Luc Guyot May 6 '19 at 23:31
• To begin with, I cannot quite understand why $G$ has the structure you say it has, being made out of those two groups you intersect. Could you please make a more precise statement about what $G$ is as a subgroup of $\mathbf{Z}[1/3]^2$? Or maybe there is something I should know in order to have a visual idea on what that intersection is! – Alex Doe May 12 '19 at 15:31
• What I'm using all the time is that the injection with dense image $\mathbf{Z}[1/p]\to\mathbf{Q}_p$ induces an isomorphism $\mathbf{Z}[1/p]/\mathbf{Z}\to\mathbf{Q}_p/\mathbf{Z}_p$. Therefore, there's a canonical bijection between the set of subgroups of $\mathbf{Z}[1/p]^d$ containing $\mathbf{Z}^d$ and the set of subgroups of $\mathbf{Q}_p^d$ containing $\mathbf{Z}_p^d$. And in the latter it is sometimes easier to define things (because we have the subspaces $V$ of $\mathbf{Q}_p^d$, so each $V+\mathbf{Z}_p^d$ defines a subgroup). – YCor May 12 '19 at 15:44