Vrftsjtryk

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 $P leq G$.

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.