Cantor-Schröder-Bernstein theorem for finitely generated abelian groups

Question

I study theory of groupa now. I tried but I can't solve this problem:

Let $A$ and $B$ be finitely generated Abelian groups, and each one of them is isomorphic to a subgroup of the other. Prove that $A\cong B$.

I reasoned like this: let A generated by $<a_1, a_2, ... , a_n>$ and B generated by $<b_1, b_2, ... , b_m>$.

Let $$f: A →f(A)$$ - isomorphism from A to a subgroup of B and $$g: B → g(B)$$ - isomorhpism from B to a subgroup of A.

We have: $A\cong f(A)\subseteq B\cong g(B)\subseteq A$. That is, image of generators of group A are contained in group B and vice versa. Am I right? Then how can I prove that $A\cong B$?

Answer 1

We will need the following result (which is usually proven in the context of the structure theorem for finitely generated abelian groups, but we don't need the full theorem here):

If $A$ is a finitely generated abelian group, then its torsion subgroup $\mathrm{Tor}(A)$ is finite. Moreover, it is a direct summand of $A$, and the quotient $A/\mathrm{Tor}(A)$ is a finitely generated free abelian group, so isomorphic to some $\mathbb{Z}^r$. Its rank $r$ is the rank of $A$, i.e. the largest size of a linearly independent subset of $A$.

Now let $A,B$ be two finitely generated abelian groups for which there are monomorphisms $A \to B$ and $B \to A$. These induce monomorphisms $\mathrm{Tor}(A) \to \mathrm{Tor}(B)$ and $\mathrm{Tor}(B) \to \mathrm{Tor}(A)$. But then these torsion groups are finite abelian groups of the same order, so any injection between them must be a bijection. So actually, $\mathrm{Tor}(A) \to \mathrm{Tor}(B)$ must be an isomorphism.

Since $A \to B$ is a monomorphism, the rank of $A$ is at most the rank of $B$. Since we also have a monomorphism $B \to A$, the ranks must coincide. Hence, $A/\mathrm{Tor}(A)$ and $B/\mathrm{Tor}(B)$ are free abelian groups of the same rank, hence isomorphic.

Therefore, $A \cong \mathrm{Tor}(A) \oplus A/\mathrm{Tor}(A) \cong \mathrm{Tor}(B) \oplus B/\mathrm{Tor}(B) \cong B$.

Remark. For general abelian groups, the statement is not correct. See SE/2461137.