Suppose to have $G$ a simple group of order n. Let p a prime dividing n. If G has more than $\frac{n}{p^2}$ conjugacy classes then prove that all p-Sylow subgroups are abelian.
I've no idea how to connect this propriety with p-Sylow. I've reasoned in that way: Suppose $G$ non abelian, then $G'=G$. $|G|=1+\sum_{\chi \in IrrG:\chi(1)>1}\chi(1)^2= 1+\sum_{\chi \in IrrG:p|\chi(1)}\chi(1)^2+\sum_{\chi \in IrrG:p \not{|}\chi(1)}\chi(1)^2\geq 1+p^2t +\sum_{\chi \in IrrG:p\not{|}\chi(1)}\chi(1)^2$.
I've taken $P\in Syl_p(G)$ and suppose $x\in P$ then $|cl_G(x)|=|N(P):C(P)||P:C_P(x)|$
Since there are more than $n/p^2$ irreducible (complex) representations, there must be a nontrivial representation of degree less than $p$, which must be faithful, because $G$ is simple.
But then on restricting to a Sylow $p$-subgroup $P$, this representation must be a sum of $1$-dimensional representations, so $P$ is abelian.