Let $G$ be a simple group of order 168, I have to show that it has at least 14 elements of order 3.
Using Sylow's theorems I proved that if $n_3$ is the number of 3-Sylows then $n_3\in\{4,7,28\}$, but now I don't know how to continue, I have to exclude the possibility of $n_3=4$, could you help me?
Reading this question Prove there is no element of order 6 in a simple group of order 168 I saw that actually $n_3=28$ could you explain me why?
Best Answer
The group $G$ acts by conjugation on the set of its own $3$-Sylow subgroups. That means that there is a homomorphism from $G$ into $S_{n_3}$, whose image is a transitive subgroup of $S_{n_3}$. Since $G$ is simple, the kernel of this map must be trivial, so $n_3!$ must be at least $168 = 2^3\times 3\times 7$.
This observation rules out $n_3=4$.