One of the infinite abelian groups that is important to infinite abelian group theory and which has shown up naturally in previous posts on infinitary fundamental groups is the Baer-Specker group, often just called the Specker group. This post isn’t all that topological, but I think it’s appropriate considering that infinite group theory is becoming increasingly important to this kind of infinite product-driven algebraic topology.
The Baer-Specker group is the countably infinite direct product of copies of the additive group of integers:
The elements* of the Specker group are sequences of integers and addition is component-wise .
We’ll write for the sequence whose only non-zero component is a in the n-th position. The countable set forms a basis for the free abelian subgroup consisting of the sequences that terminate in ‘s.
But the Specker group is uncountable (as is any countably infinite product of sets with cardinality ), which means the set cannot generate the whole group – equivalently is proper.
So is a free abelian group or not?
On the face of it, the answer is not obvious. We answered the same question (with a negative answer) for the earring group in a two-part post (Part I and Part II). This question about the Specker group is a bit easier since it is an abelian group with simpler combinatorial structure. In this post, I’ll walk through a nice, direct proof of the following theorem.
Theorem 1: is not a free group.
You’ll notice the argument is similar in spirit to the argument used to show is not free: basically, we’ll assume the group is free and exploit the infinite structure to find an integer which has infinitely many divisors – something which obviously can’t happen.
*It is also common to view the Specker group as the group of functions from the natural numbers to integers.
Apparently, Reinhold Baer  first proved Theorem 1 in 1937 (or at least stronger results that imply it). Ernst Specker  re-proved this result among a number of other original results about . So the foundational theory of is often attributed to Specker as well. The proof I’m using in this post is a nice one from the AMM .
Free abelian groups and divisibility
First, let’s review some terminology.
Recall that a basis for an abelian group is a subset such that every element can be written uniquely as a linear combination where for all but finitely many . An abelian group is free if such a basis exists.
We will use the following elementary fact about free abelian groups.
Lemma 2: Suppose is a free abelian group with basis , is a subset, and is the free subgroup generated with basis . The quotient is a free abelian group for which the set of cosets form a basis.
Also, we’ll deal with some divisibility issues.
If is an abelian group, a non-zero element, then we say a natural number divides if
for some . We call a divisor of and will be interested in cases when an element has finitely or infinitely many divisors. For example, every integer obviously has only finitely many divisors (using prime decomposition) but every non-zero rational number has infinitely many divisors since for every .
Lemma 3: Every non-zero element of a free abelian group has only finitely many divisors.
Proof. Suppose is a free abelian group with basis and is non-zero. Write uniquely using the basis. Fix some non-zero coefficient such that . Notice that if divides , then for some . But this means is equal to both
By the uniqueness of the representation of in the given basis is equal to for some .
We have just argued that whenever divides , then also divides the integer . Since integers only have finitely many divisors, can only have finitely many divisors.
The Specker group is not free
Proof of Theorem 1. Suppose is a free abelian group with basis . Since is uncountable, the indexing set must also be uncountable.
Write each element as a finite linear combination of basis elements. For fixed , let so that is the set of indices which has non-zero coefficient for some element . Notice that is countable since it is the countable union of finite sets . This means the free abelian subgroup generated by the basis is also countable.
Remark: It is worthwhile to point out that we have constructed to be the smallest free subgroup (generated by our basis) containing all of the sequences .
Here are a few important observations:
- since for each .
- Since is uncountable, the complement is also uncountable. Therefore the quotient group is uncountable.
- By Lemma 2, is a free abelian group with basis .
Definition: Let’s say a sequence is multiplicative if for each , each quotient of consecutive terms is an integer not equal to (automatically this means for any ). Let be the set of multiplicative sequences.
For example, and are multiplicative. Let’s quickly convince ourselves that there are uncountably many multiplicative sequences.
Lemma 4: There are uncountably many multiplicative sequences in .
Proof. Consider the set of sequences . Obviously this set is uncountable since, for instance, it contains the uncountable subset . Define a function by . Clearly is an injection and since has an uncountable subset, must be uncountable.
Since is uncountable and is countable, there exists some multiplicative sequence such that .
Recall that and so we have
for each .
Since is multiplicative, the element
where the first non-zero term is in the n+1-st position, is a well-defined element of the Specker group. Notice that
So in the free abelian group , we have for all showing that has infinitely many divisors in . But according to Lemma 3, this cannot happen in a free abelian group so we obtain a contradiction.
This concludes the proof that is not free.
How is all this related to the earring group?
In an earlier post, I wrote up a proof of the fact that was not a free (non-abelian) group. The argument was similar the one above for the Specker group but was a bit more sophisticated. Actually, we proved a stronger fact:
Let be the homotopy class of the loop which traverses the n-th circle of the earring space once in the counterclockwise direction.
Fact: The homomorphism to the Specker group given by is injective and has image .
Neither identifying the image of nor proving injectivity of is easy business. See this post if you’re looking for the proof.
Viewing as the infinite wedge of circles as a subspace of the infinite torus , the inclusion map induces the surjective homomorphism to the Specker group. Another way to think about this map is that the n-th component of is the winding number of around the n-th circle. For example, and .
In a similar way, let’s define a homomorphism
and prove the analogous theorem.
Theorem 5: The homomorphism given by is injective and has image . Consequently, is countable.
Proof. Since for each , the following diagram commutes:
Since is surjective, is injective. Since both and are injective, so is . Also (since the triangle commutes). For each , let , be the projection onto the n-th coordinate. Then showing that .
Theorem 5 implies Theorem 1: The fact that is countable gives another proof that the Specker group is not free since if it the Specker group is free with (necessarily uncountable) basis , then the group
of all functions is uncountable.
This viewpoint using the earring group should not be considered an easier proof of Theorem 1 since the hard work is still present on the non-abelian side. On the other hand, exploring the relationship between the two “non-freeness” results – namely that the non-abelian version implies the abelian one – is worth looking at.
 R. Baer, Abelian groups without elements of finite order, Duke Math. J. 3 (1937) 68-122.
 S. Schroer, Baer’s Result: The infinite product of the integers has no basis, The American Mathematical Monthly 115, No. 7 (2008), 660-663.
 E. Specker, Additive Gruppen von Folgen ganzer Zahlen, Port. Math. 9 (1949) 131-140.