$|Z(G) \cap G'|$ is not divisible by p if $G$ has abelian p-Sylow subgroups In Exercise §11.12 of “Methods of Representation Theory” by Curtis and Reiner, the reader is asked to prove the statement given in the title. In other words: If $P$ is an abelian $p$-Sylow subgroup of a finite group $G$, and if $x \in P \cap Z(G)$ is any non-trivial element (of order $p$, say), we have to show there is a homomorphism $\varphi \colon G \to A$ into some abelian group $A$ such that $\varphi(x) \neq 1$.
If the transfer homomorphism is known, this is an easy task. Indeed, the transfer homomorphism $\varphi \colon G \to P$ satisfies $\varphi(x) = x^{|G:P|} \neq 1$. But this morphism is only introduced later in the book. I suspect the authors want the reader to construct a linear representation $\varphi \colon G \to \mathbb{C}^\times$ instead.
Maybe a desired proof would start like this: Since $P$ is abelian, there is a linear character $\lambda$ of $P$ such that $\lambda(x) \neq 1$. Then $\chi = \lambda^G$ is a character of $G$ of degree $\chi(1) = |G : P|$ with $x \notin \operatorname{Ker}(\chi)$. More precisely, we have $\chi_{|\langle x \rangle} = |G:P| \cdot \lambda_{|\langle x \rangle}$ (since $x \in Z(G)$). In particular, we have $x \notin \operatorname{Ker}(\psi)$ for all irreducible constituents $\psi$ of $\chi$. So it suffices to show that $\chi$ has at least one linear constituent.
Any ideas how to continue from here? Note that I have not used that $P$ is a Sylow group so far. 

Added later: As Dietrich Burde's comment shows, I was actually quite close: Instead of looking for a constituent of $\chi$, one only needs to consider the determinant $\varphi = \det \chi$. This is a homomorphism $G \to \mathbb{C}^\times$ satisfying $\varphi(x) = \lambda(x)^{|G:P|} \neq 1$.
By essentially the same argument, we can also prove the more general statement: $P \cap Z(G) \cap G' \leq P'$ for all (not necessarily abelian) Sylow subgroups $P \leq G$.
 A: Here is the total proof.
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.
