# Is every torsion-free abelian group isomorphic to $\mathbb{Z}^n$?

Is every torsion-free abelian group isomorphic to $$\mathbb{Z}^n$$ for some natural $$n$$?

I think it is true just for finitely generated abelian torsion-free groups...

• Consider $\mathbb{Q}$. You are right, it needs to be finitely generated. Commented Mar 9, 2018 at 12:29
• Every abelian group that is isomorphic to $\mathbb{Z}^n$ is finitely generated. Your question is basically "are there any non-finitely generated abelian groups?" Commented Mar 9, 2018 at 13:12

Every finitely-generated torsionfree abelian group is isomorphic to $$\mathbb Z^n$$ for some $$n$$, but the proof crucially relies on the finiteness hypothesis (as most of them use induction, one way or another).

In fact, there are torsionfree abelian groups that are not isomorphic to $$\mathbb Z^{(I)}$$ for any $$I$$, or to $$\mathbb Z^I$$ for any $$I$$ (the two possible generalizations one may think of for infinite $$I$$).

As stated in the comments, $$\mathbb Q$$ is one such example, but so is the group of $$p$$-adic integers $$\mathbb Z_p$$ or the additive group of the ring $$\mathbb Z[1/p]$$.

In $$\mathbb Z^{(I)}$$ or $$\mathbb Z^I$$, there are no elements that are divisible by arbitrarily high integers, which is what fails in these examples.

(If you know homological algebra, here's some food for thought : over $$\mathbb Z$$, torsion-free modules are precisely the flat modules; and modules of the form $$\mathbb Z^{(I)}$$ are precisely the projective modules. In these terms, your question amounts to "over $$\mathbb Z$$, does flat imply projective ?")

You have all the elements to try $\mathbb{Z}^\mathbb{N}$ with componentwise addition.

To further clarify the question and the mentioned comments and in order to provide a complete answer:

An elementary argument shows that the following proposition holds:

Proposition. Every finitely-generated torsion-free abelian group is isomorphic to $$\mathbb{Z}^{n}$$ for some $$n\geq 1.$$

proof. Suppose that $$G$$ is generated by $$\{a_{1},a_{2},\ldots,a_{n}\};$$ i.e., we have:
$$G:=\Big\{\sum_{i=1}^{n}k_{i}a_{i}: k_{i}\in\mathbb{Z}\Big\}.$$

Let $$\bar{\mathbf{1}}_{i}$$ be the element in $$\mathbb{Z}^{n}$$ all of whose components are equal to zero except for the $$i$$th component that equals $$1.$$ It is easy to verify that the function that maps each $$a_{i}$$ to $$\bar{\mathbf{1}}_{i}$$ is a group isomorphism.$$\qquad\blacksquare$$

And, as mentioned in the comments, the additive group of the rationals is an example of a torsion-free abelian group that is not finitely-generated.

Looking at $$\langle\mathbb{Q},+,0\rangle$$ from another point of view, it is an ordered divisible group. It can be shown that an abelian group is torsion-free if and only if it is orderable (some hints are available in the link below about an argument that uses the proposition above). Finally, an interesting question might be the following:

Question. Is there an abelian divisible torsioned group?

More detailed explanations can be found in the following question:
Is there a natural example of a divisible torsioned (= periodic) abelian group?

• Your "elementary argument" is wrong. $\mathbb Z$ is generated by $\{2,3\}$ for instance, but certainly $\mathbb Z$ is not isomorphic to $\mathbb Z^2$ Commented Jul 27, 2019 at 13:19
• @Max: What you have said is not a counter example for the given argument since $\mathbb{Z}$ is in fact isomorphic to itself! This misunderstanding is resolved by considering a suitable language for the theory of groups. I will explain more in the following comments: Commented Jul 28, 2019 at 6:38
• 1- If you look more precisely at the definition of a finitely-generated group in my post, the coefficients are ranging over $\mathbb{Z}$ and not over $\mathbb{N}.$ Hence, $\mathbb{Z}$ in generated by $1.$ In first-order logic, you can consider different languages to study a theory, for example the group theory. The question of which language to choose depends on your perspective and actually your purpose of studying that theory whether it be decidability issues, complexity of formulas, and ... But for a general purpose, people usually choose a language that is natural to that theory. Commented Jul 28, 2019 at 6:43
• 2- Now, if you let the language of group theory, $\mathcal{L}_{gr}$ contain, beside the binary operator of the group, distinguished symbols for the ''identity element" and the "inverse map", then every $\mathcal{L}_{gr}$-substructure becomes a subgroup which is counted as a benefit for this language. It is an elementary exercise to check that without having these symbols, a substructure might not be a subgroup because it would not be necessarily closed under taking inverses. Commented Jul 28, 2019 at 6:52
• User A : it is not a counterexample to the claim (because the claim is correct !), it is, however, a counterexample to the proof. There is no misunderstanding on my part, only an erroneous proof - using irrelevant notions of model theory will not change that. Commented Jul 28, 2019 at 9:07