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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.