## A Free Abelian Group which Is not Free

In this post, we provide an example of an abelian group which is not free, i.e. does not have a basis. Namely, we show that the additive group formed by the countably infinite direct product of the integer group $(\mathbb{Z},+)$, which we denote by $\mathbb{Z}^{\mathbb{N}}$ is not free. Here, $\mathbb{N}$ is the set of natural numbers, including zero.

Definition 1. An abelian group $F$ is said to be free abelian if $F$ is isomorphic to the direct sum (note, not direct product) of infinite cyclic groups. Explicitly, there exists a subset $X \subset F$ consisting of elements of infinite order, which we call a basis of $F$, with $F \cong \bigoplus_{x \in X} \langle{x}\rangle$, where $\langle{x}\rangle$ denotes the subgroup generated by $x$.

We allow the possibility that $X = \emptyset$, in which case $F$ is the trivial group.

An equivalent characterization of a free abelian group $F$ is that $F$ has a basis such that every element $g \in F$ can be uniquely written as an expression of the form

$\begin{displaystyle} g = \sum_{x \in F} m_{x}x, \end{displaystyle}$

where almost all (i.e. all but finitely many) $m_{x} = 0$.

Let $\mathbb{Z}^{\mathbb{N}}$ denote the countably infinite direct product (note, not direct sum) of $\mathbb{Z}$.

Theorem 2. Let $\varphi: \mathbb{Z}^{\mathbb{N}} \rightarrow \mathbb{Z}$ be a homomorphism. Then there exists $m \in \mathbb{N}$ such that

$\begin{displaystyle} \varphi\left(\left\{(a_{j})_{j=0}^{\infty} \in \mathbb{Z}^{\mathbb{N}} : a_{j} = 0 \ \forall j < m \right\}\right) = \left\{0\right\} \end{displaystyle}$

Proof. If $\varphi = 0$, then we’re done; so assume otherwise. For $k \geq 1$, define a subset $U_{k} \subset \mathbb{Z}^{\mathbb{N}}$ by

$\begin{displaystyle} U_{k} := \left\{(a_{j})_{j=0}^{\infty} \in \mathbb{Z}^{\mathbb{N}} : a_{j} = 0 \ \forall j < k\right\} \end{displaystyle}$

Let $\left\{b_{k} : k \in \mathbb{N}\right\}$ be an enumeration of $\mathbb{Z}$. Set $a^{(0)} := 0$ and $r_{0} := 1$. Suppose we have chosen $a^{(0)},\cdots,a^{(n)} \in \mathbb{Z}^{\mathbb{N}}$ and $r_{0},\cdots,r_{n} \in \mathbb{Z}$ such that

$\begin{displaystyle} \varphi\left(a^{(k)} + r_{0}\cdots r_{k} U_{k+1}\right) \cap \left\{b_{k}\right\} = \emptyset, \indent \forall 0 \leq k \leq n \end{displaystyle}$

If $\varphi(U_{n+1}) = \left\{0\right\}$, then we’re done. Otherwise, we can choose

$\begin{displaystyle} a^{(n+1)} \in a^{(n)} + r_{0}\cdots r_{n}U_{n+1} \text{ such that }\varphi(a^{(n+1)}) \neq b_{n} \end{displaystyle}$

Choose $r_{n+1} \in \mathbb{Z}$ such that $0 < \left|\varphi(a(n+1))- b_{k}\right| < r_{k+1}$ and therefore $b_{n} \notin \varphi(a^{(n+1)}) + r_{n+1}\mathbb{Z}$. Hence, for any $c = (c_{j})_{j=1}^{\infty} \in \mathbb{Z}^{\mathbb{N}}$,

$\begin{displaystyle} \varphi(a^{(n+1)}+ r_{n+1}c) = \varphi(a^{(n+1)}) + \underbrace{r_{n+1}\varphi(c)}_{\in r_{n+1}\mathbb{Z}} \neq b_{n}, \end{displaystyle}$

which implies that $\varphi\left(a^{(n+1)} + r_{n+1}\mathbb{Z}^{\mathbb{N}}\right) \cap \left\{b_{n}\right\} = \emptyset$. Either this process terminates after finitely many steps, or by induction, we obtain an element $a = (a_{n}^{(n)})_{n=1}^{\infty} \in \mathbb{Z}^{\mathbb{N}}$. Observe that by construction,

$\begin{displaystyle} a \in a_{n} + r_{0}\cdots r_{n}U_{n+1} \Longrightarrow \varphi(a) \neq b_{n}, \indent \forall n \in \mathbb{N} \end{displaystyle}$

But since $\left\{b_{n} : n \in \mathbb{N}\right\} = \mathbb{Z}$, we conclude that $\varphi(a) \notin \mathbb{Z}$, which is a contradiction. $\Box$

The reader can verify that the following result is a ready consequence of the preceding lemma.

Proposition 3. For $n \in \mathbb{N}$, let $j_{n}: \mathbb{Z} \rightarrow \mathbb{Z}^{\mathbb{N}}$ be the $n^{th}$ inclusion

$\begin{displaystyle} j_{n}(a) := (\underbrace{0,\cdots,0}_{j-1},a,0,\cdots), \indent \forall a \in \mathbb{Z} \end{displaystyle}$

If we have a countable collection $\left\{\psi_{n} : \mathbb{Z} \rightarrow \mathbb{Z}\right\}_{n \in \mathbb{N}}$ of homomorphisms, which contains an infinite subet of nonzero maps, then there does not exist a homomorphism $\varphi: \mathbb{Z}^{\mathbb{N}} \rightarrow \mathbb{Z}$ such that

$\begin{displaystyle}\varphi \circ j_{n} = \psi_{n} \end{displaystyle}$,     $\begin{displaystyle} \forall n \in \mathbb{N} \end{displaystyle}$

As an application of the preceeding proposition, taking $\psi_{n}: \mathbb{Z} \rightarrow \mathbb{Z}$ to be the identity mapping for all $n \in \mathbb{N}$, we see that there does not exist a homomorphism $\varphi: \mathbb{Z}^{\mathbb{N}} \rightarrow \mathbb{Z}$ such that

$\begin{displaystyle} \varphi \circ j_{n}(m) = m \end{displaystyle}$,     $\begin{displaystyle} \forall n \in \mathbb{N} \end{displaystyle}$

Theorem 4. $\mathbb{Z}^{\mathbb{N}}$ is not a free abelian group.

Proof. Suppose $\mathbb{Z}^{\mathbb{N}}$ is free abelian and therefore there exists a $\mathbb{Z}$-module isomorphism

$\begin{displaystyle}\Phi: \mathbb{Z}^{\mathbb{N}} \rightarrow \bigoplus_{x \in X} \langle{x}\rangle \end{displaystyle}$

where $X \subset \mathbb{Z}^{\mathbb{N}}$ consists of elements of infinite order. Let $\left\{x_{n} : n \in \mathbb{N}\right\}$ be an at most countable subset of $X$, and for each $n \in \mathbb{N}$, let $\psi: \bigoplus_{x \in X}\langle{x}\rangle \rightarrow \mathbb{Z}$ be the projection map

$\begin{displaystyle} \psi\left(\sum_{x}m_{x}x\right) := \sum_{x} m_{x}, \indent \forall z = \sum_{x}m_{x}x \in \bigoplus_{x \in X}\langle{x}\rangle \end{displaystyle}$

By composition, we obtain a homomorphism $\varphi:= \psi \circ \Phi: \mathbb{Z}^{\mathbb{N}} \rightarrow \mathbb{Z}$. By Lemma 2, there exists $m \in \mathbb{Z}^{\geq 1}$ such that $\varphi(U_{m}) = \left\{0\right\}$, where $U_{m}$ is defined as above. $\Box$