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.
Best Answer
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.