# On free abelian groups

I'm learning about the concept of a free abelian group. First question: nowhere it is stated that these groups cannot be finite, but the definition seems to imply it. Is this true?

Second question: an isomorphism to a free abelian group $$A = \left< S \right> = \left< \{s_i\}_{i=1}^n \right>$$ is given as

$$\begin{eqnarray*} &ℤ^n &\tilde\longrightarrow A \\ &(c_i)_{i=1}^n &\mapsto \sum_{s ∈ S} c_ss \end{eqnarray*}$$

Then a claim is made that I don't understand:

"The induced isomorphism $$A/2A \cong (ℤ/2ℤ)^n$$ now shows that $$A/2A$$ is of order $$2^n$$, from which we conclude that the rank of $$A$$ does not depend on the choice of a basis for $$A$$"

What is this induced isomorphism? Is it induced by the (first) Isomorphism Theorem? What exactly is $$A/2A$$, and why are we looking at it? How is the final conclusion drawn?

The trivial group is free abelian (of rank $$0$$), and finite. But every nontrivial free abelian group is infinite: given an element $$s$$ of the free basis, you can map $$s$$ to the generator of the infinite cyclic group $$C_{\infty}$$ and ever other element of the basis to the identity. The universal property gives you a homomorphism from the free abelian group to this infinite group, so the free abelian group has an infinite quotient, hence is itself infinite.
Under the original map, the subgroup $$2A$$ corresponds to the subgroup $$(2\mathbb{Z})^n$$. If you have a morphism $$f\colon G\to K$$, a normal subgroup $$M\triangleleft K$$, and a normal subgroup $$N\triangleleft G$$ such that $$f(N)\subseteq M$$, then you get an “induced map” $$(G/N)\to (K/M)$$ given by $$gN\mapsto f(g)M$$. In the case at hand, $$f$$ is an isomorphism, $$2A$$ corresponds to $$(2\mathbb{Z})^n$$, so you actually get an isomorphism between the quotients.
$$A/2A$$ is just the quotient of $$A$$ modulo the “even” elements. It gives you an abelian group such that $$2a=0$$ for all $$a$$; this makes it into a vector space over the field of $$2$$ elements, and we know lots of things about such vector spaces (in particular, that they have a well-defined dimension). So we are looking at it to leverage that knowledge into showing that the size of a basis of a free abelian group is unique.
• I'm almost there. (The universal property argument is a bit too abstract for me.) Also, I see that $A/2A$ has cardinality $2^n$, but I don't see that this translates to information about $A$ itself. Commented Jan 16, 2020 at 22:07
• @JosvanNieuwman: Suppose $A$ had a basis of size $m$; the same argument tells you that $A/2A$ has order $2^m$. But the size of $A/2A$ does not depend on the basis. So $2^n=2^m$ which means $n=m$. That is, any two bases for $A$ have the same size. Commented Jan 16, 2020 at 22:12
• @JosvanNieuwman If you don’t want the universal property, then simply note that the elements with $c_t=0$ and $c_s$ an arbitrary integer give you infinitely many distinct elements of $A$. Commented Jan 16, 2020 at 22:14
• Wow. I literally forgot that $n$ was the cardinality of the basis.. I should go to bed. Thanks a bundle! Commented Jan 16, 2020 at 22:20