## A criterion for the existence of unique group homomorphisms from an abelian group

Let $A = \langle x_1 \rangle \times \cdots \langle x_t \rangle$ be a finite abelian group with $|x_i| = n_i$ for each $i$.

1. Find a presentation for $A$.
2. Prove that if $G$ is a group containing commuting elements $g_1,\ldots,g_t$ such that $g_i^{n_i} = 1$ for each $i$, then there is a unique group homomorphism $\theta : A \rightarrow G$ such that $\theta(x_i) = g_i$.

1. We claim that $A \cong B = \langle a_1, \ldots, a_t \ |\ a_i^{n_i} = 1, a_ia_j = a_ja_i \rangle$, where $1 \leq i,j \leq t$. Indeed, the “standard basis” elements $e_i$, consisting of $x_i$ in the $i$-th coordinate and 1 in all other coordinates, satisfy these relations.
2. Note that every element of $A$ can be written uniquely as $(x_i^{b_i})_{i=1}^t$. Define $\theta((x_i^{b_i})) = \prod_{i=1}^t g_i^{b_i}$, where the product on the right hand side is as computed inside $G$. It is clear that $\theta(x_i) = g_i$ for each $i$. Moreover, $\theta$ is a homomorphism because the $g_i$ commute pairwise.

Finally, suppose $\sigma$ is a homomorphism $A \rightarrow G$ such that $\sigma(x_i) = g_i$ for each $i$. Then $\sigma((x_i^{b_i})) = \prod \sigma(x_i)^{b_i}$ $= \prod \theta(x_i)^{b_i}$ $= \theta((x_i^{b_i}))$. Thus $\sigma = \theta$ (products computed inside $G$), so that $\theta$ is unique.