### Introduction

Assuming the axiom of choice, any vector space possesses the pleasant but prosaic property* that it is determined up to isomorphism by the cardinality of its basis.

For instance, consider $ \prod_\omega \mathbb{Z}/2$ and $ \oplus_{2^\omega} \mathbb{Z}/2$. Both are vector spaces over the finite field $ \mathbb{Z}/2$ so to show that they are isomorphic, we need to show that their respective bases have the same cardinality. The vector space on the right is written as a direct summand and so we can see that its basis must have size $ \mathfrak{c}$. On the other hand, the vector space $ \prod_\omega \mathbb{Z}/2$ has cardinality $ \mathfrak{c}$ over a finite field, so its basis must have the same cardinality as the space itself. Aren't vector spaces a walk in the park; a piece of cake; easy as pie (ok, enough metaphors?!)?

### From $ \mathbb{Z}/2$ to $ \mathbb{Z}$ Modules

But what if we sent the above proof to a publisher who didn't yet have the "2" character or the "/" installed on her printing press? Then all hell would break loose because $ \prod_\omega \mathbb{Z}$ and $ \oplus_{2^\omega} \mathbb{Z}$ aren't vector spaces any more, and the previous paragraph would be rife with errors. But they certainly are abelian groups, and they have a bit more spice than those vector spaces. So are they isomorphic? They do have the same cardinality. Fortunately for us, Baer ("Abelian Groups without Elements of Finite Order", Duke Math J. 3 (1937), pp. 88-122) answered this question in the negative (fortunately, because otherwise abelian groups would be less exciting). In fact, this question is particularly interesting to me because I had wondered about it a few months ago, and now I have the answer, thanks to Faith's book for the references.

Now note that $ \oplus_{2^\omega}\mathbb{Z}$ is free by definition, so in order to show that $ \prod_{\omega}\mathbb{Z}\not\cong\oplus_{2^\omega}\mathbb{Z}$, we just have to show that $ \prod_{\omega}\mathbb{Z}$ is not free. Now, some countable torsion-free abelian groups like $ \mathbb{Q}$ will never trick us into thinking that they are free. In fact, $ \mathbb{Q}$ is not even projective! (Though it is flat and we all know that a module is flat iff it is a direct limit of projectives.)

But $ \prod_{\omega}\mathbb{Z}$ is a bit more sinister. There's no need to worry, however, because the proof is not too complicated, and can be found in the Kaplansky classic, "Infinite Abelian Groups". In fact, I'll give a brief sketch of this proof, but I encourage the reader to go through the full proof.

### Sketch of Proof of Non-Isomorphism

The sketch goes as follows: let $ A =\prod_{\omega}\mathbb{Z}$ for brevity. Now, if $ A$ cannot be countably generated as a $ \mathbb{Z}$ module, because if it were, then $ A$ would also be countable, which is false. Write an element in the infinite product as $ (a_1,a_2,\dots)$, and let $ B$ be the submodule of $ A$ consisting of the set of all sequences $ (a_1,a_2,\dots)$ such that the power of $ 2$ dividing $ a_i$ goes to infinity as $ i$ goes to infinity. The map $ (a_1,a_2,a_3,\dots)\mapsto (2a_1,4a_2,8a_3,\dots)$ is an injective homomorphism $ A\hookrightarrow B$.

If $ A$ is free then so is $ B$ because $ \mathbb{Z}$ is a principal ideal domain so if $ A$ is uncountably generated then so is $ B$. But then $ B/pB$ as a vector space over $ \mathbb{Z}/2$ has a countable basis, which is impossible since as a module $ B$ has an uncountable basis.

So we see that $ \prod_{\omega}\mathbb{Z}\not\cong\oplus_{2^\omega}\mathbb{Z}$.

Note that the above proof goes through essentially without modification if $ \mathbb{Z}$ is replaced by any principal ideal domain that is NOT a field.

### But there are Subgroups!

Even though $ \mathbb{Z}^\omega$ is not free, the subset of all bounded sequences of $ \mathbb{Z}^\omega$ is in fact free. This is a result of Specker, and according to A. Zastrow ("The non-abelian Specker group is free", Journal of Algebra 229, 55-85 (2000)), Specker actually first proved his result using the continuum hypothesis although later Nöbeling discovered a proof that avoided the CH.

For more on Specker groups and related properties, I encourage the reader to peruse Fuch's two-volume series, "Infinite Abelian Groups" or some of the papers mentioned above.

### Concluding Remarks

Infinite abelian groups can be tricky!

*Alliteration rules!