Finitely generated abelian groups are elementarily equivalent iff they are isomorphic

Statement

Suppose $G$ and $H$ are Finitely generated abelian group (?)s. Then, $G$ and $H$ are Elementarily equivalent groups (?) if and only if they are Isomorphic groups (?).

Facts used

1. Quotients of elementarily equivalent abelian groups by multiples of n are elementarily equivalent
2. Structure theorem for finitely generated abelian groups
3. Finite groups are elementarily equivalent iff they are isomorphic
4. Torsion subgroups of elementary equivalent abelian groups are elementarily equivalent

Proof

By fact (1), we have that, for every $n$, $G/nG$ and $H/nH$ are elementarily equivalent. By fact (2), we have that:

$G \cong \mathbb{Z}^r \oplus G_1$

where $G_1$ is a finite abelian group.

$H \cong \mathbb{Z}^s \oplus H_1$

where $H_1$ is a finite abelian group.

First, choose $n$ as an integer greater than $1$ that is a multiple of the exponents of both $G_1$ and $H_1$. Then, $G/nG$ and $H/nH$ are elementarily equivalent, and these are:

$G/nG \cong (\mathbb{Z}/n\mathbb{Z})^r, \qquad H/nH \cong (\mathbb{Z}/n\mathbb{Z})^s$.

These are both finite groups, so by fact (3), they are isomorphic, so $r = s$.

Further, by fact (4), the torsion subgroups of $G$ and $H$ are elementarily equivalent, so, since they are finite, fact (3) yields that $G_1 \cong H_1$.

Thus, $G \cong H$.