Let $G$ be a finite abelian subgroup of $(\mathbb{C}^{\times})^n$, which is a group with elements $(a_1,\ldots,a_n)\in \mathbb{C}^n$ and multiplication is done by
$$(a_1,\ldots,a_n) \ast (b_1,\ldots,b_n) = (a_1b_1, \ldots, a_nb_n)$$
Question: Show that there exists a finite abelian subgroup of $(\mathbb{C}^{\times})^{n-1}$ (call it $G'$) and a positive integer $m$ such that $$G \cong (\mathbb{Z}/m\mathbb{Z}) \times G'$$
Here is my progress: in $G$, there exists a term with maximal order. In a finite abelian group, given two elements $a,b$, we can construct an element in $\langle a,b\rangle$ with order $\text{lcm}(\text{ord}(a),\text{ord}(b))$. We can show if $(a_1,\ldots,a_n)\in G$ of maximal order, say $m$, then
$$G=\langle (a_1,\ldots,a_n) \rangle \times G/\langle (a_1,\ldots,a_n) \rangle \cong \mathbb{Z}/m\mathbb{Z} \times G/\langle (a_1,\ldots,a_n) \rangle$$
So we just need $G/\langle (a_1,\ldots,a_n) \rangle$ to be isomorphic to a finite abelian subgroup of $(\mathbb{C}^{\times})^{n-1}$. One possible idea is to construct an element from each $\langle (a_1,\ldots,a_n) \rangle$-coset where the order of that element in $G$ is equal to the order of that element's coset in $G/\langle (a_1,\ldots,a_n) \rangle$. Call this element the representative of $G/\langle (a_1,\ldots,a_n) \rangle$. Furthermore, there exists $1\le i\le n$ such that for all $g\in G$, the representative of $g\langle (a_1,\ldots,a_n)\rangle$ has the $i$th coordinate equal to $1$.
I'm not sure if it suffices to construct $(a_1,\ldots,a_n)\in G$ such that $(a_1,\ldots,a_n)$ have the maximum order in $G$ and among all elements $(g_1,\ldots,g_n)\in G$, $\text{ord}(g_1)\mid \text{ord}(a_1)$. I think I know how to construct such an element.