Squeeze Theorem for Abelian Groups

Solution 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 $C$.