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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ Jan 16 '20 at 22:07
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    $\begingroup$ @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. $\endgroup$ Jan 16 '20 at 22:12
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    $\begingroup$ @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 $A$. $\endgroup$ Jan 16 '20 at 22:14
  • $\begingroup$ Wow. I literally forgot that $n$ was the cardinality of the basis.. I should go to bed. Thanks a bundle! $\endgroup$ Jan 16 '20 at 22:20

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