Let $G$ be a finite simple group and $\chi$ be an irreducible character of $G$ of degree $p$ (prime). Prove that any $p$-sylow subgroup of $G$ is of size $p$.
I have no idea how to solve the problem. There is also a small hint to the question.
Hint: if $P$ is a non abelian Sylow-$p$ subgroup then $Z(P) \subseteq Z(\chi)$.
$Z(\chi)=\{\,g\in G : |\chi(g)|=|\chi(1)|\,\}$
I didn't get how to use the hint to my advantage.
I started with the argument that $G$ cannot be a $p$-group since a simple $p$-group has to be abelian in which case it will be abelian of prime order and hence all irreducible characters are of degree $1$. Hence $|G|=p^a. m$ (with $m>1$). But I don't get how to proceed further.
Since $G$ is simple, we must have $Z(\chi)=1$, as $Z(\chi)$ is always a normal subgroup. So by Theorem 3.8 in Isaacs's book, $\chi$ vanishes on $Z(P)\setminus \{1\}$, and $\chi(1)=p$ by assumption. It follows that $Z(P)$ has order $p$. Let $1\neq x\in Z(P)$, and $D\colon G\to \operatorname{GL}(p,\mathbb{C})$ a representation affording $\chi$. Then $D(x)$ is a matrix of order $p$ and has trace $\chi(x)=0$. It follows that the eigenvalues of $D(x)$ are exactly the different $p$-th roots of unity (including $1$), each with multiplicity $1$. Then the centralizer of $D(x)$ in $\operatorname{GL}(p,\mathbb{C})$ is abelian, and thus $P$ is abelian. Thus $P=Z(P)$ has order $p$.