Torsion-free commutative groups of a given cardinality

Question

If an abelian group $G$ is torsion-free, it has at least one subgroup isomorphic to $\Bbb{Z}$, given by $⟨g⟩ := \{ g^n | n \in \Bbb{Z} \}$, for any non-identity element $g$. This subgroup is obviously countable. I'm curious if there is a minimal, torsion-free, uncountable abelian group $G_1$ such that, for every torsion-free abelian group $A$ of uncountable cardinality (i.e. $|A| \geq \aleph_1$), there is an embedding of $G_1$ into $A$?

Answer 1

As pointed out in comments, it's not clear what you mean by "minimal". But there is, up to isomorphism, a unique uncountable abelian group $G_1$ that embeds in every uncountable torsion-free abelian group. As suggested in comments by @Robert Shore, this is the free abelian group of rank $\aleph_1$.

Suppose $A$ is an uncountable torsion-free abelian group. Then, since $A$ is torsion-free, the natural map $A\to A\otimes_\mathbb{Z}\mathbb{Q}$ is injective, and so $A\otimes_\mathbb{Z}\mathbb{Q}$ is also uncountable. But $\dim_\mathbb{Q}(A\otimes_\mathbb{Z}\mathbb{Q})$ is one definition of the (torsion-free) rank of $A$, so $\operatorname{rank}A$ is uncountable.

But another equivalent definition of $\operatorname{rank}A$ is the maximal rank of a free abelian subgroup of $A$, so $A$ has a free abelian subgroup of uncountable rank, which in turn must have a free abelian subgroup of rank $\aleph_1$.

There is no other group $G_1$ with this property, since, taking $A$ to be free abelian of uncountable rank, $G_1$ would have to be free abelian, since every subgroup of a free abelian group is free abelian.

Answer 2

Jeremy Rickard beat me to it, but here is a slightly lower-tech answer (though maybe the same if you unpack everything).

Claim 1: Suppose $A$ is a torsion-free abelian group and $B \le A$ is a subgroup such that $A / B$ is a torsion group. Then $|A| = |B|$.

Proof: If $B$ is finite then $A$ and $B$ are trivial and the claim holds, so assume $B$ is infinite. For every $a \in A$ there is some $n_a < \infty$ such that $a^{n_a} \in B$. Suppose $n_{a_1} = n_{a_2} = n$ and $a_1^n = a_2^n$. Then $(a_1a_2^{-1})^n = 1$, so $a_1 = a_2$ since $A$ is torsion-free. It follows that the map $A \to \mathbb N \times B$ defined by $a \mapsto (n_a, a^{n_a})$ is injective, so $|A| \le \aleph_1 |B| = |B| \le |A|$, so $|A| = |B|$. $\square$

For a set $X$ let $\mathbb Z^X$ denote the free abelian group with basis $X$, i.e., the direct sum of $|X|$-many copies of $\mathbb Z$.

Claim 2: (AC) Let $A$ be an uncountable torsion-free abelian group. Then there is an injective homomorphism $\mathbb Z^{|A|} \to A$. In particular there is a subgroup of $A$ isomorphic to $\mathbb Z^{\aleph_1}$.

Proof: Let $\alpha$ be the first ordinal of cardinality $|A|$ and consider the set of all injective homomorphisms $\mathbb Z^\gamma \to A$ where $\gamma \le \alpha$. We order this set by extension, i.e., if $f_i : \mathbb Z^{\gamma_i} \to A$ ($i=1,2$) then $f_1 \le f_2$ means $\gamma_1 \le \gamma_2$ and $f_2 | \mathbb Z^{\gamma_1} = f_1$. An application of Zorn's lemma shows that there is a maximal element, say $f : \mathbb Z^\gamma \to A$. We claim that $\gamma = \alpha$. Otherwise, let $B = f(\mathbb Z^\gamma)$ and note that $|B| = \aleph_1 |\gamma| < |A|$, so the previous claim shows that $A / B$ is not torsion-free. Let $a \in A$ be an element of infinite order modulo $B$. Then we can use $a$ to extend $f$ to an injective homomorphism $\mathbb Z^{\gamma+1} \to A$, in contradiction to the maximality of $f$. $\square$