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Is there any chance to find a torsion-free, non-cyclic, abelian group $A$ such that $\operatorname{Aut}(A)=\mathbb Z_2$? ($\mathbb Z_2$ is the cyclic group of order $2$)

Notation

$\operatorname{Aut}(A)$ is the group of autmorphisms of $A$.

darij grinberg
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W4cc0
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  • How about the abelian group $G=\langle a,b\rangle={a^ib^j;|;(i,j)\in\Bbb Z^2}$? The automorphisms $G\to G$ are ${1,\phi}$ where $\phi(a)=b,\phi(b)=a$ so that $\phi^2=1$. $G$ is clearly torsion free and not cyclic since it is generated by 2 elements. – Yong Hao Ng Dec 31 '13 at 16:55
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    @YongHaoNg: the automorphism group of your $G$ is infinite. For instance $\phi(a) = a$, $\phi(b) = a^i b$ has order infinity for every $i \neq 0$. – Jack Schmidt Dec 31 '13 at 16:57
  • @JackSchmidt Ah I see! Thanks. =D – Yong Hao Ng Dec 31 '13 at 17:00

1 Answers1

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Yes. Let $A=\left\{ \tfrac ab : a,b \in \mathbb{Z}, b \textrm{ is square-free} \right\}$.

If $A \leq \mathbb{Q}$ then its ring of endomorphisms is a subring of $\mathbb{Q}$. Its automorphisms are the invertible elements, those $\tfrac ab \in \mathbb{Q}^\times$ such that multiplication by them is both injective (automatic) and surjective (special). An abelian group $A$ is said to be $n$-divisible for an element $n \in \mathbb{Z}\setminus\{0\}$ when $nA = A$, that is, when multiplication by $n$ is surjective. Suppose $\tfrac ab A = A$ with $\gcd(a,b)=1$. Find $u,v$ such that $au+bv=1$, then for $x \in A$, we have $\tfrac 1b x = \tfrac{au+bv}{b} = u \tfrac ab x + v x \in A$, so $A$ is $b$-divisible. By symmetry, since $\tfrac ba = {\left(\tfrac ab\right)}^{-1} \in \newcommand{\Aut}{\operatorname{Aut}}\Aut(A)$ as well, $A$ is $a$-divisible.

In other words, $\Aut(A) = \left\{ \tfrac ab \in \mathbb{Q}^\times : A \text{ is $a$-divisible and $b$-divisible } \right\}$.

Every abelian group is $1$-divisible and $-1$-divisible, so $Z_2 \cong \{1,-1\} \leq \Aut(A)$. Most torsion-free rank 1 groups are only $1$ and $-1$ divisible, but they might also be $p$-divisible for some primes $p$. If they are, then their automorphism group is bigger. Explicitly, $\Aut(A) \cong Z_2 \times F$, where $F$ is the free abelian group with basis those primes $p$ such that $A$ is $p$-divisible.

For the specific group $A=\left\{ \tfrac ab : a,b \in \mathbb{Z}, b \textrm{ is square-free} \right\}$ that I mention, $A$ is not $p$-divisible for any prime $p$ (since the denominators would eventually be divisible by $p^2$).

Most torsion-free rank-one $A$ work, you just need to make sure they aren't $p$-divisible for any prime $p$.

Your earlier question ($\Aut(A)$ has a unique element of order 2) was satisfied by all torsion-free rank-one groups.

Jack Schmidt
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