Let $G$ be a finite solvable group, where every Sylow subgroup is abelian. I want to show that if $A\lhd G$ is an abelian normal subgroup, then $$ A=(A\cap Z(G))(A\cap G')$$
This is easy if $A$ is a minimal normal subgroup, so I've tried to induct on the $G$-composition length of $A$. So if $L<A$ is a normal subgroup of $G$, maximal in the sense that there are no normal subgroups of $G$ strictly between $L$ and $A$, I want to show either $A\le LZ(G)$ or $A\le LG'$.
If there exists $a\in A$ and $g\in G$ with $[g,a]\not\in L$, then $$ A= L(A\cap G')= A\cap LG'$$ so that case is done.
The other case is where $[G,A]\le L$. Now I need to find an $a\in A\setminus L$ that is central. But I have no idea how to do that.
Since $A$ is the direct product of its Sylow subgroups, each of which is normal in $G$, we can suppose that $A$ is a $p$-group for some prime $p$.
Since $G$ has abelian Sylow subgroups, the group $C \cong G/C_G(A)$ of automorphisms of $A$ that is induced by conjugation by $G$ is a a $p'$-group, and it is enough to prove that there is an $a \in A \setminus L$ that is central in the semidirect product $A \rtimes C$.
Let $b \in A \setminus L$. Then, since $[G,A] = [C,A] \le L$, we have $C^b \le CL$. Since $C$ is a $p'$-group, we can apply the Schur-Zassenhaus Theorem to deduce that $C$ and $C^b$ are conjugate in $CL$, and hence by an element $l \in L$. So $bl^{-1} \in A \cap N_G(C) \le C_G(C)$, and so $a = bl^{-1}$ is central.