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

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