# 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 (?).

## 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, $G \cong H$.