Let $G$ be a group such that $G'$ abelian and any abelian normal subgroup of $G$ is finite. Show that $G$ is finite.
The centralizer of $G'$ in $G$ has finite index in $G$, so we can assume that it is the whole of $G$: i.e. $G' \le Z(G)$. Let $A=A_1$ be an abelian normal subgroup of $G$ containing $Z(G)$. So $A$ is finite, and hence its centralizer $C_G(A)$ has finite index in $G$. So if $G$ is infinite, then $A$ is properly contained in $C_G(A)$, and then choosing $g \in C_G(A) \setminus A$, we get a larger abelian normal subgroup $A_2 = \langle A,g \rangle$. (It is normal because it contains $G'$.) Hence, if $G$ is infinite, then we can construct an infinite strictly ascending chain $A_1 < A_2 < A_3 < \cdots$ of abelian normal subgroups of $G$, and its union is an infinite abelian normal subgroup, contradiction.