Martin Isaacs's exercise 3.7 (character theory of finite groups)

Question

I would need some help with this exercise:

Let $\chi\in{Irr(G)}$ be faithful, and suppose $\chi(1)=p^a$ for some prime p. Let $P\in{Syl_{p}(G)}$, and suppose that $C_{G}(P)\nsubseteq{P}$. Show that $G'\lt{G}$.

Thanks a lot in advance.

Answer 1

Hint: Use theorem 3.8 to get a non-kernel element of $Z(\chi)$ and show that $\det(\chi)(x) \neq 1$.

Sketch of first part:

Use problem 2.16 on a $p'$-subgroup of $C_G(P)$ to ensure that we don't just always have $\chi = 0$.

Here is a proof:

For every $x \in C_G(P)$, one has $P \subseteq C_G(x)$ and so $|\mathcal{K}(x)| =[G:C_G(x)]$ divides $[G:P]$ which is prime to $p^a=\chi(1)$. Hence $\chi(x)=0$ or $x \in Z(\chi)$. Let $Q \subseteq C_G(P)$ be a $p'$-group (which exists by hypothesis). If $\chi $ vanished on $Q-\{1\}$, then $|Q|$ would divide $\chi(1)$ by problem 2.16, but this is impossible. Hence there is some $x \in Q-\{1\}$ with $x \in Z(\chi)$. Now consider $(\det \chi)(x)$ and the representation $X$ affording $\chi $. Since $x \in Z(\chi)$, we have $X(x) = \lambda I$, where $\lambda$ is a $|x|$th root of unity (and $|x| \gt 1$ is prime to $p$). Hence $(\det \chi)(x) = \det(X(x)) = \lambda^{p^a}$. This can only be $1$ if $\lambda = 1$, but this is impossible since $x$ is not in the kernel of $\chi$. Hence $(\det\chi)(x)$ is not $1$, and $\det\chi \neq 1_G$ is a nonprincipal linear character, giving that $[G:G'] \gt 1$. $\square$