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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.