A lifting property for finite abelian groups

Posted on 2010-06-17 by nbloomf
Tags: abelian-group, finite-group, homomorphism, group-presentation, generating-set, d-and-f

Let \(A\) be the finite abelian group \[A = \mathbb{Z}/(n_1) \times \mathbb{Z}/(n_2) \times \cdots \times \mathbb{Z}/(n_t).\]

If \(G\) is a group containing elements \(g_1, g_2, \ldots, g_t\) such that \(g_i^{n_i} = 1\) for each \(i\) and \(g_ig_j = g_jg_i\) for all \(i\) and \(j\), then there is a unique group homomorphism \(\theta : A \rightarrow G\) such that \(\theta(1_i) = g_i\) for each \(i\). (Here, \(1_i\) is the element of \(A\) with a \(\underline{1}\) in the \(i\)th entry and \(\underline{0}\) elsewhere.)

Let \(\theta \subseteq A \times G\) consist of the pairs \[\left(\left(\underline{b_i}\right)_{i=1}^t, \prod_{i=1}^t g_i^{b_i}\right).\] We claim that \(\theta\) is a group homomorphism. Clearly \(\theta\) is total.

To see that \(\theta\) is well-defined, suppose we have tuples \(\left(\underline{b_i}\right), \left(\underline{c_i}\right) \in A\) such that \((\underline{b_i}) = (\underline{c_i})\); that is, for each \(i\) we have \(b_i \equiv c_i \pmod{n_i}\), and thus \(b_i - c_i = k_in_i\) for some \(k_i\). Now \[\prod_{i=1}^t g_i^{b_i - c_i} = \prod_{i=1}^t g_i^{k_in_i} = \prod_{i=1}^t (g_i^{n_i})^{k_i} = \prod_{i=1}^t 1^{k_i} = \prod_{i=1}^t 1 = 1,\] and thus (since the \(g_i\) commute pairwise) \[\prod_{i=1}^t g_i^{b_i} = \prod_{i=1}^t g_i^{c_i}.\] Thus \(\theta\) is a mapping.

Now note that if \(\left(\underline{b_i}\right), \left(\underline{c_i}\right) \in A\), we have \[\begin{array}{rcl} \theta\left(\left(\underline{b_i}\right)+\left(\underline{c_i}\right)\right) & = & \theta\left(\underline{b_i+c_i}\right) \\ & = & \prod_{i=1}^t g_i^{b_i + c_i} \\ & = & \prod_{i=1}^t g_i^{b_i} g_i^{c_i} \\ & = & \prod_{i=1}^t g_i^{b_i} \cdot \prod_{i=1}^t g_i^{c_i} \\ & = & \theta\left(\underline{b_i}\right) \cdot \theta\left(\underline{c_i}\right). \\ \end{array}\]

(From line 3 to 4 we use the fact that the \(g_i\) commute pairwise.) So \(\theta\) is a group homomorphism. Finally, it is clear that \(\theta(1_i) = g_i\) for each \(i\).

Now suppose \(\sigma : A \rightarrow G\) is a group homomorphism such that \(\sigma(1_i) = g_i\) for each \(i\). We have \[\begin{array}{rcl} \sigma\left(\underline{b_i}\right) & = & \sigma\left(\sum_{i=1}^t b_i 1_i\right) \\ & = & \prod_{i=1}^t \sigma(1_i)^{b_i} \\ & = & \prod_{i=1}^t g_i^{b_i} \\ & = & \theta\left(\underline{b_i}\right) \\ \end{array}\] so that \(\sigma = \theta\) as claimed.