# How to show that, in this case, all the Sylow $p$-subgroups of $G$ are abelian.

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?

• Nice problem, but are you sure this is true? Where does this question come from? – Nicky Hekster Feb 5 at 10:40
• @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. – ciccio Feb 5 at 10:48

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.
• 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. – ciccio Feb 13 at 13:23
• 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. – Nicky Hekster Feb 13 at 19:03
• 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 $p$. – Nicky Hekster Feb 13 at 19:07