Characterization of free abelian groups of finite rank

Let S be a set. The group presented by (S, R) where R = \{ [s,t] \ |\ s,t \in S \} is called the free abelian group on S. Denote it by A(S). Prove that A(S) has the following universal property: If G is any abelian group and \varphi S \rightarrow G is a set function, then there is a unique group homomorphism \Phi : A(S) \rightarrow G such that \Phi|_S = \varphi. Deduce that if A is a free abelian group on a set of (finite) cardinality n, then A \cong \mathbb{Z}^n.


Recall from the universal property of free groups that there is a unique group homomorphism \Psi : F(S) \rightarrow G such that \Psi|_S = \varphi. Moreover, clearly R \leq \mathsf{ker}\ \Psi. Then there is a unique group homomorphism \Psi : A(S) \rightarrow G such that \Psi = \Phi \circ \pi. Identifying s in S with sR in A(S), we have \Phi|_S = \varphi. This gives existence.

To see uniqueness, note that any homomorphism \Phi : A(S) \rightarrow G extending \varphi is uniquely determined by \Psi, which is uniquely determined by \varphi.

Now let n \in \mathbb{N} and let S = \{1, \ldots, n \}. If n = 0, then A(S) is a quotient of F(S) = 1, so that in fact A(S) = 1. Suppose n > 0, and for each 1 \leq k \leq n, define \varphi(k) to be the n-tuple with 1 in the kth coordinate and 0 in all other coordinates. By the universal property, there exists a unique group homomorphism \Phi : A(S) \rightarrow \mathbb{Z}^n extending \varphi; \Phi is surjective since \Phi[S] generates \mathbb{Z}^n. Now if w = s_1^{e_1} \cdots s_n^{e_n} \in \mathsf{ker}\ \Phi, note that since A(S) is abelian, we can collect terms so that s_i = i. Now \Phi(w) = (e_1, \ldots, e_n) = (0, \ldots, 0), and in fact w = 1. Thus \Phi is injective, and we have A(S) \cong \mathbb{Z}^n.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: