Character theory - exercise 3.4 from Isaacs Let $G$ be a simple group and suppose $\chi\in Irr(G)$ with $\chi(1)=p$ a prime, Show that a Sylow $p-$subgroup of $G$ has order $p$.
The indication provided is: if the Sylow $p-$subgroup $P$ is nonabelian, then $Z(P) \subset Z(\chi)$ where $Z(\chi)$ is the center of the character $\chi$: $
Z(\chi)=\{g\in G: |\chi(g)|=\chi(1)\}$. 
I am trying to prove the indication. 
Once proven. If $P$ is nonabelian, then $Z(P) \subset Z(\chi)$ . But since $Z(\chi)\triangleleft G$ and $G$ is simple, $Z(\chi)=1$ or $G$. But $p=\chi(1) | [G:Z(\chi)]$ and so necessarily $Z(\chi)=1$, contradicting $Z(P)>1$ as $P$ is a $p-$group. 
If $P$ is abelian, let $x\in P$. Then $P\le C_G(x)$ and so $|Conj_G(x)|=[G:C_G(x)]\wedge \chi(1)=1$. By Burnside's theorem, $\chi(x)=0$ if $x\notin Z(\chi)=1$. Hence $\chi$ vanishes on $P-\{1\}$. By a classic result (problem 2.16), $[P:1]$ divides $\chi(1)=p$, which gives the result. 
 A: Any reference here is to Isaac's book Character Theory of Finite Groups; let me write out the proof more in Isaacs' style. Some preparatory remarks. Let $P \in Syl_p(G)$ ($P$ exists and is non-trivial, since $\chi(1)=p \mid |G|$). Now $\chi_P=\sum a_{\theta}\theta$ for integers $a_{\theta} \gt 0$ and $\theta \in Irr(P)$. Since $\theta(1) \mid |P|$, each $\theta(1)$ is a power of $p$. Clearly, $\chi_P(1)=\chi(1)=p=\sum a_{\theta}\theta(1) \geq \theta(1)$ for each of these irreducible constituents $\theta$ of $\chi_P$. Hence, $\theta(1) \in \{1,p\}$. Further, observe that $\chi$ is non-linear, so $G$ cannot be abelian. Since $G$ is simple, $Z(G)=1$, and the non-linearity of $\chi$ gives $Z(\chi) \lt G$, whence $Z(\chi)=1$ and so is $ker(\chi)=1$ ($\chi$ is faithful). Now we have two cases.
(1) If $\theta(1)=p$ for some $\theta$, then $\chi_P=\theta$ (of course, the corresponding $a_{\theta}=1$). Hence $\chi_P$ is irreducible.  But then $ker(\chi_P)=P \cap ker(\chi)=1$, so (Lemma 2.27(f)) $Z(P)\subseteq Z(\chi_P)=P \cap Z(\chi)=1$, contradicting $P$ being a non-trivial $p$-group. Hence: 
(2) $\theta(1)=1$ for all irreducible constituents $\theta$ of $\chi_P$, that is, $\chi_P$ is the sum of some linear characters. But then, (Lemma (2.21) and (2.23)(a)), $1=ker(\chi_P)=\bigcap_{\theta} ker(\theta) \supseteq P'$. We conclude that $P$ is abelian.
Note that we have proved sofar that if $G$ is a simple group having an irreducible character of degree $p$, then a $p$-subgroup must be abelian! The proof can now be finished by applying Theorem(3.13) and it is here that we are using that $P$ is a Sylow $p$-subgroup.
Note (added February 4th 2019) A similar line of reasoning as above gives the following. Let $G$ be a simple group and suppose $\chi\in Irr(G)$ with $\chi(1)=p$ a prime. Let $q$ be another prime, with $q \mid |G|$ and $q \gt p$. Then a $q$-subgroup of $G$ is abelian. In particular a Sylow $q$-subgroup is abelian.
