Squeeze Theorem for Abelian Groups

Question

Recently I saw this theorem :

Assume that $A\leq B\leq C$ with $A\cong\mathbb{Z}^n\cong C$. Then $B\cong\mathbb{Z}^n$.

I wonder whether it is possible to generalise this result to arbitrary finitely generated abelian groups.

What I mean is this : If $A\leq B\leq C$ are finitely generated abelian groups with $A\cong C$ then $B\cong C$.

Using the fundamental theorem of finitely generated abelian groups it seems like it is true but I could not prove it.

Edit : If we remove the condition that $C$ is abelian then we know that the claim is not true for free groups. There are subgroups of $F_2$ which is not isomorphic to $F_2$ but contain a subgroup which is isomorphic to $F_2$. But I also do not know an example of a solvable group $C$ with there exists $A\leq B\leq C, A\cong C$ but $B\not\cong C$.

Answer 1

For the case of solvable group, let $C$ be the wreath product of two copies of the integers. So this is a split extension of a direct sum of countably many copies $Z_i$ of the integers ($i$ running over the integers), by a copy of the integers, spanned by $g$, say, and written multiplicatively, with $g$ conjugating $Z_i$ to $Z_{i+1}$. This is a $2$-generated metabelian, hence solvable, group.

Now let $B$ be the subgroup generated by the $Z_i$ and $g^2$. Clearly it is not isomorphic to $C$, as there are two orbits of the $Z_i$ under the span of $g^ 2$, corresponding to $i$ even and $i$ odd. It is a $3$-generated group.

Finally, let $A$ be the subgroup of $B$ generated by the $Z_i$, for $i$ even, and $g^ 2$. This is isomorphic to $C$.