The Baer-Specker Group

One of the infinite abelian groups that is important to infinite abelian group theory and which has shown up naturally in previous posts on wild 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 wild 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 (a_n) of integers a_n\in\mathbb{Z} and addition is component-wise (a_n)+(b_n)=(a_n+b_n).

We’ll write s_n=(0,0,..,0,1,0,0,...) for the sequence whose only non-zero component is a 1 in the n-th position. The countable set \{s_n|n\geq 1\} forms a basis for the free abelian subgroup \bigoplus_{n=1}^{\infty}\mathbb{Z} consisting of the sequences (a_1,a_2,...,a_n,0,0,...) that terminate in 0‘s.

But the Specker group \prod_{n=1}^{\infty}\mathbb{Z} is uncountable (as is any countably infinite product of sets with cardinality >2), which means the set \{e_n|n\geq 1\} cannot generate the whole group – equivalently \bigoplus_{n=1}^{\infty}\mathbb{Z} is proper.

So is \displaystyle\prod_{n=1}^{\infty}\mathbb{Z} 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 Hawaiian earring group \pi_1(\mathbb{H}) 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 simple, direct proof of the following theorem.

Theorem 1: \displaystyle\prod_{n=1}^{\infty}\mathbb{Z} is not a free group.

You’ll notice the argument is similar in spirit to the argument used to show \pi_1(\mathbb{H}) 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 \mathbb{Z}^{\mathbb{N}} of functions f:\mathbb{N}=\{1,2,3,...\}\to\mathbb{Z} from the natural numbers to integers.


Apparently, Reinhold Baer [1] first proved Theorem 1 in 1937 (or at least stronger results that imply it). Ernst Specker [3] re-proved this result among a number of other original results about \prod_{n=1}^{\infty}\mathbb{Z}. So the foundational theory of \prod_{n=1}^{\infty}\mathbb{Z} is often attributed to Specker as well. The proof I’m using in this post is a nice one from the AMM [2].

Free abelian groups and divisibility

First, let’s review some terminology.

Recall that a basis for an abelian group A is a subset \{e_{\lambda}|\lambda\in\Lambda\}\subset A such that every element a\in A can be written uniquely as a linear combination a=\sum_{\lambda}n_{\lambda}e_{\lambda} where n_{\lambda}=0 for all but finitely many \lambda. An abelian group A is free if such a basis exists.

We will use the following elementary fact about free abelian groups.

Lemma 2: Suppose A is a free abelian group with basis \{e_{\lambda}|\lambda\in\Lambda\}, J\subset\Lambda is a subset, and H=\langle e_{\lambda}|\lambda\in J\rangle is the free subgroup generated with basis \{e_{\lambda}|\lambda\in J\}. The quotient A/H is a free abelian group for which the set of cosets \{e_{\lambda}+H|\lambda\in\Lambda\backslash J\} form a basis.

Also, we’ll deal with some divisibility issues.

If A is an abelian group, a\in A a non-zero element, then we say a natural number n divides a if

\displaystyle a=nx=\underbrace{x+x+...+x}_\text{n terms}

 for some x\in A. We call n a divisor of a and will be interested in cases when an element has finitely or infinitely many divisors. For example, every integer a\in \mathbb{Z} obviously has only finitely many divisors (using prime decomposition) but every non-zero rational number \frac{p}{q}\in \mathbb{Q} has infinitely many divisors since \frac{p}{q}=n\left(\frac{p}{nq}\right) for every n\geq 1.

Lemma 3: Every non-zero element of a free abelian group has only finitely many divisors.

Proof. Suppose A is a free abelian group with basis \{e_{\lambda}|\lambda\in\Lambda\} and a\in A is non-zero. Write a=\sum_{\lambda}n_{\lambda}e_{\lambda} uniquely using the basis. Fix some non-zero coefficient \lambda_0\in\Lambda such that n_{\lambda_0}\neq 0. Notice that if n divides a, then a=nx_n for some x_n=\sum_{\lambda}m_{\lambda}e_{\lambda}. But this means a is equal to both

nx_n=\sum_{\lambda}(nm_{\lambda})e_{\lambda} and \sum_{\lambda}n_{\lambda}e_{\lambda}.

By the uniqueness of the representation of a in the given basis n_{\lambda_0} is equal to nm_{\lambda} for some \lambda.

We have just argued that whenever n divides a, then n also divides the integer n_{\lambda_0}. Since integers only have finitely many divisors, a can only have finitely many divisors. \square

The Specker group is not free

Proof of Theorem 1. Suppose G=\prod_{n=1}^{\infty}\mathbb{Z} is a free abelian group with basis \{e_{\lambda}|\lambda\in\Lambda\}. Since G is uncountable, the indexing set \Lambda must also be uncountable.

Write each element s_n=(0,0,...,0,1,0,0,...) as a finite linear combination s_n=\sum_{\lambda}m_{\lambda,n}e_{\lambda} of basis elements. For fixed n, let B(n)=\{\lambda\in\Lambda|m_{\lambda,n}\neq 0\} so that B=\bigcup_{n=1}^{\infty}B(n) is the set of indices \lambda which has non-zero coefficient for some element s_n. Notice that B is countable since it is the countable union of finite sets B(n). This means the free abelian subgroup H=\langle e_{\lambda}|\lambda\in B\rangle generated by the basis \{e_{\lambda}|\lambda\in B\} is also countable.

Remark: It is worthwhile to point out that we have constructed H to be the smallest free subgroup (generated by our basis) containing all of the sequences s_n.

Here are a few important observations:

  1. \bigoplus_{n=1}^{\infty}\mathbb{Z}\subseteq H since s_n\in H for each n\geq 1.
  2. Since \Lambda is uncountable, the complement \Lambda\backslash B is also uncountable. Therefore the quotient group G/H is uncountable.
  3. By Lemma 2, G/H is a free abelian group with basis \{e_{\lambda}+H|\lambda\in\Lambda\backslash B\}.

Definition: Let’s say a sequence (a_n)\in G is multiplicative if for each n\geq 1, each quotient \displaystyle\frac{a_{n+1}}{a_n} of consecutive terms is an integer not equal to \pm 1 (automatically this means a_n\neq 0 for any n). Let M be the set of multiplicative sequences.

For example, (n!)=(1!,2!,3!,4!,...) and (2^n)=(2,4,8,16,...) are multiplicative. Let’s quickly convince ourselves that there are uncountably many multiplicative sequences.

Lemma 4: There are uncountably many multiplicative sequences in G.

Proof. Consider the set of sequences S=\{(b_n)\in G||b_n|>1\text{ for each }n\}. Obviously this set is uncountable since, for instance, it contains the uncountable subset \prod_{n=1}^{\infty}\{2,3\}=\{(b_n)\in G|b_n\in\{2,3\}\}. Define a function f:S\to M by f(b_n)=(b_1,b_1b_2,b_1b_2b_3,...,b_1b_2\cdots b_k,...). Clearly f is an injection and since M has an uncountable subset, M must be uncountable. \square

Since M is uncountable and there are only countably many cosets of H (which partition G, there exists some multiplicative sequence a=(a_n)\in M such that a \notin H.

Recall that \bigoplus_{n=1}^{\infty}\mathbb{Z}\subseteq H and so we have

h_n=\sum_{k=1}^{n}a_ks_k=(a_1,a_2,....,a_n,0,0,...)\in H  for each  n\geq 1.

Since a=(a_n) is multiplicative, the element

\displaystyle x_n=\left(0,0,...,0,\frac{a_{n+1}}{a_n},\frac{a_{n+2}}{a_n},\frac{a_{n+3}}{a_n},...\right)

– where the first non-zero term is in the n+1-st position – is a well-defined element of the Specker group. Notice that

a_nx_n=(0,0,...,0,a_{n+1},a_{n+2},...)=a-h_n\in a+H.

So in the free abelian group G/H, we have a+H=a_n(x_n+H) showing that a+H has infinitely many divisors a_1,a_2,a_3,... in G/H. But according to Lemma 3, this cannot happen in a free abelian group so we obtain a contradiction.

This concludes the proof that G=\prod_{n=1}^{\infty}\mathbb{Z} is not free. \square

How is all this related to the Hawaiian earring group?

In an earlier post, I wrote up a proof of the fact that \pi_1(\mathbb{H}) 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 g_n=[\ell_n]\in\pi_1(\mathbb{H}) be the homotopy class of the loop \ell_n which traverses the n-th circle of the Hawaiian earring once in the counterclockwise direction.

Fact: The homomorphism \eta:Hom(\pi_1(\mathbb{H}),\mathbb{Z})\to\prod_{n=1}^{\infty}\mathbb{Z} to the Specker group given by \eta(f)=(f(g_n))=(f(g_1),f(g_2),f(g_3),...) is injective and has image \bigoplus_{n=1}^{\infty}\mathbb{Z}.

Neither identifying the image of \eta nor proving injectivity of \eta is easy business. See this post if you’re looking for the proof.

Viewing \mathbb{H} as the infinite wedge \bigvee_{n=1}^{\infty}S^1 of circles as a subspace of the infinite torus \prod_{n=1}^{\infty}S^1, the inclusion map \bigvee_{n=1}^{\infty}S^1\to\prod_{n=1}^{\infty}S^1 induces the surjective homomorphism \epsilon:\pi_1(\mathbb{H})\to\pi_1\left(\prod_{n=1}^{\infty}S^1\right)=\prod_{n=1}^{\infty}\mathbb{Z} to the Specker group. Another way to think about this map is that the n-th component of \epsilon([\alpha]) is the winding number of \alpha around the n-th circle. For example, \epsilon(g_n)=s_n and \epsilon([\ell_1\ell_2\ell_3...])=(1,1,1,...).

In a similar way, let’s define a homomorphism

\zeta:Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\to\prod_{n=1}^{\infty}\mathbb{Z}, by \zeta(f)=(f(s_n))=(f(s_1),f(s_2),...)

and prove the analogous theorem.

Theorem 5: The homomorphism \zeta:Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\to\prod_{n=1}^{\infty}\mathbb{Z} given by \eta(f)=(f(g_n))=(f(s_1),f(s_2),f(s_3),...) is injective and has image \bigoplus_{n=1}^{\infty}\mathbb{Z}. Consequently, Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\cong\bigoplus_{n=1}^{\infty}\mathbb{Z} is countable.

Proof. Since \epsilon(g_n)=s_n for each n\geq 1, the following diagram commutes:

Since \epsilon is surjective, Hom(\epsilon,\mathbb{Z}) is injective. Since both \eta and Hom(\epsilon,\mathbb{Z}) are injective, so is \zeta. Also Im(\zeta)\subseteq Im(\eta)=\bigoplus_{n=1}^{\infty}\mathbb{Z} (since the triangle commutes). For each n\geq 1, let f_n:\prod_{n=1}^{\infty}\mathbb{Z}\to\mathbb{Z}, f_n((a_n))=a_n be the projection onto the n-th coordinate. Then \zeta(f_n)=s_n showing that Im(\zeta)=\bigoplus_{n=1}^{\infty}\mathbb{Z}. \square

Theorem 5 implies Theorem 1: The fact that Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)\cong\bigoplus_{n=1}^{\infty}\mathbb{Z} is countable gives another proof that the Specker group is not free since if it the Specker group is free with (necessarily uncountable) basis B, then the group

Hom\left(\prod_{n=1}^{\infty}\mathbb{Z},\mathbb{Z}\right)=Hom\left(\bigoplus_{b\in B}\mathbb{Z},\mathbb{Z}\right)\cong Set(B,\mathbb{Z})

of all functions B\to\mathbb{Z} is uncountable.

This viewpoint using the Hawaiian earring 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.


[1] R. Baer, Abelian groups without elements of finite order, Duke Math. J. 3 (1937) 68-122.

[2] S. Schroer, Baer’s Result: The infinite product of the integers has no basis, The American Mathematical Monthly 115, No. 7 (2008), 660-663.

[3] E. Specker, Additive Gruppen von Folgen ganzer Zahlen, Port. Math. 9 (1949) 131-140.

I found Specker’s paper freely available here but disappointingly, couldn’t get access to Baer’s paper without using my University library.
This entry was posted in Baer-Specker group, Free abelian groups, Free groups, Group homomorphisms, Hawaiian earring, Infinite Group Theory and tagged , , , , , , . Bookmark the permalink.

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