# Finitely generated abelian groups with the same finite quotients

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.

• 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). Commented Apr 30 at 21:11
• Tangential remark: I feel like this should be true for general residually finite groups. Commented Apr 30 at 23:15
• @LukasHeger: see mathoverflow.net/q/90885/1446 Commented May 1 at 0:01
• @SteveD interesting, thanks! Commented May 1 at 0:03

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 $$s\leq r$$).