Let $G$ be a finite solvable group, all of whose Sylow subgroups are abelian. Prove that $Z(G) \cap G' = 1$. 
Let $G$ be a finite solvable group, all of whose Sylow subgroups are abelian. Prove that $Z(G) \cap G' = 1$.

Attempt: From the second isomorphism theorem $\frac{G'}{Z(G)\cap G'} \simeq \frac{G'Z(G)}{Z(G)}$ so it suffices to show that $\frac{G'Z(G)}{Z(G)} \simeq G'$. I couldn't proceed any further though and I can't figure out how to use the fact that every Sylow subgroup is abelian. Any hint is appreciated.
Thanks.
 A: If the result is false, then there is a prime $p$ dividing $|Z(G) \cap G'|$. Let $P \in {\rm Syl}_p(G)$. Then an element of order $p$ in $Z(G) \cap G'$ lies in $P$, so $Z(G) \cap G' \cap P \ne 1$.
Consider the transfer homomorphism $\tau:G \to P/P'$ which, since $P$ is abelian, can be taken to be $\tau:G \to P$.
Now it is easy to see from the definition of $\tau$ that, for $g \in Z(G)$, we have $\tau(g) = g^k$, where $k := |G:P|$ is coprime to $p$. So $\tau$ restricted to $Z(G) \cap P$ is injective,but on the other hand ${\rm Im}(\tau)$ is abelian, so if $g \in G'$ then $\tau(g)=1$.
So $Z(G) \cap P \cap G' = 1$, contrary to assumption.
A: Just for fun, here is a different proof based on character theory of groups. The transfer map is most of the time reflected in the determinant of (a representation affording) an irreducible character.
Theorem Let $P \in Syl_p(G)$, then $G' \cap Z(G) \cap P \subseteq P'$.
Proof Choose a linear character $\lambda \in Irr(P)$ and $g \in G' \cap Z(G) \cap P$. We are going to show that $\lambda(g)=1$. Since $P'=\bigcap \{ker(\lambda) : \lambda \in Irr(P), \lambda(1)=1\}$, this will guarantee the result. Now, $\lambda^G=\sum a_{\chi}\chi$, for certain $\chi \in Irr(G)$ and $a_{\chi}$ positive integers. Since $\lambda^G(1)=|G:P|=\sum a_{\chi}\chi(1)$ is a not divisible by $p$, there must be a constituent $\chi$ of $\lambda^G$ with $p \nmid \chi(1)$. Let $\mathfrak{X}$ be a representation affording this character $\chi$. Since $g \in Z(G)$, $\mathfrak{X}(g)=\omega I$ for certain $\omega \in \mathbb{C}^*$. Note that the order of $\omega$ (in $\mathbb{C}^*$) is a divisor of the order of $g$, which is a power of $p$, since $g \in P$. But $g \in G'$, so $det(\mathfrak{X}(g))=det(\omega I)=\omega^{\chi(1)}=1$. So the order of $\omega$ divides also $\chi(1)$. We conclude that $\omega=1$, and hence $g \in ker(\mathfrak{X})=ker(\chi)$. By Frobenius Reciprocity, $\chi_P=a_{\chi}\lambda + \cdots$, whence $ker(\chi_P)=P \cap ker(\chi) \subseteq ker(\lambda)$. So $g \in ker(\lambda)$ and we are done.
Your assertion, that all $P$ are abelian, now implies that no prime divides the order of the normal subgroup $G' \cap Z(G)$, so it must be trivial. Note that we do not need solvability of the group and that the condition that all Sylow-subgroups are abelian does not imply that $G$ is solvable.
