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Let $\Gamma$ and $\Delta$ be two finitely generated abelian groups. Therefore, by the classification theorem of finitely generated abelian groups, we can assume that $\Gamma \cong \mathbb{Z}^r \oplus T_1$ and $\Delta \cong \mathbb{Z}^s \oplus T_2$, where $T_1$ and $T_2$ are finite abelian groups.

I am assuming that $\Gamma$ and $\Delta$ have the same finite quotients and I want to prove that $\Gamma \cong \Delta$.

To do this, first of all, I want to prove that $r=s$. I want to procede in the following way: if by contradiction $r>s$, then I can choose a large prime $p$ such that $p$ does not divide $|T_1||T_2|$, and construct a finite quotient $(\mathbb{Z}/p\mathbb{Z})^r$ that cannot be a quotient of $\Delta$. How can I prove that $(\mathbb{Z}/p\mathbb{Z})^r$ is not a quotient of $\Delta$? I am thankful for any hint.

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    $\begingroup$ Any image of $\mathbb{Z}^s$ can be generated by at most $s$ elements, From the theory of vector spaces, we know any generating set for $(\mathbb{Z}/p\mathbb{Z})^r$ has at least $r$ elements. You can use this to show the desired equality directly (without an argument by contradiction that never uses the hypothesis except to contradict the last line of the argument). $\endgroup$ Commented Apr 30 at 21:11
  • $\begingroup$ Tangential remark: I feel like this should be true for general residually finite groups. $\endgroup$ Commented Apr 30 at 23:15
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    $\begingroup$ @LukasHeger: see mathoverflow.net/q/90885/1446 $\endgroup$
    – Steve D
    Commented May 1 at 0:01
  • $\begingroup$ @SteveD interesting, thanks! $\endgroup$ Commented May 1 at 0:03

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The group $(\mathbb{Z}/p\mathbb{Z})^r$ is in fact a vector space over $\mathbb{Z}/p\mathbb{Z}$, of dimension $r$. From the theory of vector spaces, we know that any generating set has at least $r$ elements.

As you do, find a prime $p$ such that $pT_1=T_1$ and $pT_2=T_2$, which always exists. If $G$ is a finite $p$-group and $\pi\colon \Gamma\to G$ is a group homomorphism, then $p\Gamma\leq\ker(\pi)$, so the image is a quotient of $(\mathbb{Z}/p\mathbb{Z})^r$. Thus, if $(\mathbb{Z}/p\mathbb{Z})^k$ is a quotient of $\Gamma$, then $k\leq r$, and every value of $k$, $0\leq k\leq r$, occurs.

Symmetrically, if $(\mathbb{Z}p\mathbb{Z})^k$ is a quotient of $\Delta$, then $0\leq k\leq s$ and all such values occur.

Since we are assuming that $\Gamma$ and $\Delta$ have the same set of finite quotients, it follows that $s=r$ (from the first, we obtain that $r\leq s$, and symmetrically that $s\leq r$).

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