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Let $G$ be a finite, simple group of order $n$. Let $p$ be a prime divisor of $|G|$ and suppose that the number of conjugacy classes of $G$ is $> \frac{n}{p^2}$. Then all the Sylow $p$-subgroups of $G$ are abelian.

Clearly we may assume that $G$ is not abelian. Moreover we may assume that the power of $p$ which divides the order of $G$ is $>2$, because a group of order $p^2$ is abelian. By our assumption and by simplicity of $G$, we have $Z(G)=\{1\}$. I would like to apply in some way the Theorem of Burnside (Theorem 3.8 of Isaacs' Character theory of finite groups), but i don't know how to do it. Do you have any hints?

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  • $\begingroup$ Nice problem, but are you sure this is true? Where does this question come from? $\endgroup$ – Nicky Hekster Feb 5 at 10:40
  • $\begingroup$ @NickyHekster thank you for answering. In fact i don't know if the assertion is true. It is an exercise taken from a written test. $\endgroup$ – ciccio Feb 5 at 10:48
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Here is an answer, and also credits go to prof. Derek Holt, with which I had a mail contact about whether or not your post is true.

Let $G$ be simple, $P$ a $p$-subgroup. Assume that $P$ is not abelian (so in particular $P$ is non-trivial). Since every group of which the order divides $p^2$ is abelian, we can assume that $|P| \geq p^3$, hence $p^3 \mid |G|$. So $\frac{|G|}{p^2}$ is a positive integer, and the condition on the number of conjugacy classes $k(G) \gt \frac{|G|}{p^2}$ is equivalent to $k(G) \geq \frac{|G|}{p^2} +1$.

Let $\chi \in Irr(G)$ be a non-principal character. The irreducible constituents of $\chi_P$ cannot be all linear: if $\chi_P=\sum a_{\lambda}\lambda$ for certain linear $\lambda \in Irr(P)$, then for these $\lambda$’s, $P’ \subseteq \bigcap_{\lambda} ker(\lambda)=ker(\chi_P)=ker(\chi) \cap P= 1 \cap P=1$, since $\chi$ is faithful. This implies $P$ being abelian, contradicting our assumption. So, $\chi_P$ must have a non-linear irreducible constituent of the $p$-subgroup $P$, yielding $\chi(1) \geq p$ for all non-principal irreducible characters.

To finish the proof we are using the formula $|G|=\sum_{\chi \in Irr(G)} \chi(1)^2$. Since $G$ has only one linear character we get $|G| \geq 1 + (k(G)-1)p^2 \geq 1 + \frac{|G|}{p^2} \cdot p^2 = 1+|G|$, a contradiction. So any $p$-subgroup of $G$ is abelian and in particular its Sylow $p$-subgroups must be abelian.

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  • $\begingroup$ Great. I guess that $G$ must have one linear character because we are assuming that $G$ is not abelian (we would have had $G'=\{1\}$ if $G$ had more than one linear character ). Thank you very much, also to Prof. Derek. $\endgroup$ – ciccio Feb 13 at 13:23
  • $\begingroup$ Since you can assume $G$ is non-abelian simple, $G=G'$, so the number of linear characters is $|G:G'|=1$. Also every non-principal character must be faithful. $\endgroup$ – Nicky Hekster Feb 13 at 19:03
  • $\begingroup$ All right. Thanks! $\endgroup$ – ciccio Feb 13 at 19:06
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    $\begingroup$ Perhaps you want to try solve the following exercise: let $G$ be a simple group and $\chi \in Irr(G)$, with $\chi(1)=p$. Prove that a Sylow $p$-subgroup of $G$ is cyclic of order $p$. $\endgroup$ – Nicky Hekster Feb 13 at 19:07

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