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

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

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  • $\begingroup$ (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 $A$. $\endgroup$ – Luc Guyot May 6 '19 at 23:31
  • $\begingroup$ 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! $\endgroup$ – Alex Doe May 12 '19 at 15:31
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    $\begingroup$ 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). $\endgroup$ – YCor May 12 '19 at 15:44

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