A question in finite group theory with proof using representation theory

Question

This is not a homework.

I am a beginner in studying the representation theory and character theory and I am doing some exercises in this field, but I have a problem with an exercise:

I want to prove that for a simple group $G$ with $p \mid |G|$, where $p$ is a prime number, if the number of conjugacy classes of $G$ exceeds $\frac{|G|}{p^2}$, then the Sylow $p$-subgroups of $G$ are abelian.

I know that I can write $|G|$ as the sum of the squares of the degrees of the irreducible $\mathbb{C}$-representations and the degree of each irreducible representation divide $|G|$, but I don't know how I can correlate this fact to $p$-sylow subgroups of $G$. Maybe, I don't know some facts about the sylow subgroups of a finite group.

I will be so appreciate for any helpful answer.

Answer 1

The fact that there are more than $n/p^2$ irreducible characters whose squares sum to $|G|$ implies that at least one nontrivial character $\chi$ has degree less than $p$. Since $G$ is simple, $\chi$ is faithful, so it restricts to a faithful character $\chi_P$ of a Sylow $p$-subgroup $P$ of $G$. The degrees of the irreducible constituents of $\chi_P$ all divide $|P|$, so they must all have degree $1$. So the representation corresponding to $\chi_P$ is similar to a representation whose images are diagonal matrices, and hence $P$ must be abelian.

Answer 2

Just some observations (not a complete answer):

First, if $p^3\nmid |G|$, then the Sylow-$p$'s have order $p$ or $p^2$ and are thus abelian. So assume $p^3\mid |G|$.

If I recall correctly, you can write $|G|$ as a sum of $n$ squares, where $n$ is the number of conjugacy classes in $G$. We are given that $n>\dfrac{|G|}{p^2}$, so $|G|$ is the sum of over $\dfrac{|G|}{p^2}$ squares, one of which is $1^2$, corresponding to the trivial representation.