# A lifting property for finite abelian groups

Posted on 2010-06-17 by nbloomf

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.