Finitely generated abelian groups are elementarily equivalent iff they are isomorphic

Statement
Suppose $$G$$ and $$H$$ are fact about::finitely generated abelian groups. Then, $$G$$ and $$H$$ are fact about::elementarily equivalent groups if and only if they are fact about::isomorphic groups.

Facts used

 * 1) uses::Quotients of elementarily equivalent abelian groups by multiples of n are elementarily equivalent
 * 2) uses::Structure theorem for finitely generated abelian groups
 * 3) uses::Finite groups are elementarily equivalent iff they are isomorphic
 * 4) uses::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, $$G \cong H$$.