= Solution
A <finite nonabelian simple group> cannot be a <p-group>: a nontrivial finite <p-group> has nontrivial <centre of a group>, since its <class equation> makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a <normal subgroup>, so must be the whole group if $G$ is a <simple group>, making it abelian; an abelian <simple group> has prime order. Therefore every <Sylow subgroup> here is nontrivial and proper. It cannot be normal, so $n_p>1$.
To obtain a <simple group embedding from Sylow conjugation>, the <conjugation action on Sylow subgroups> gives a <group homomorphism> $\rho:G\to S_{n_p}$. This <group action> is transitive by the <Sylow theorems> and is nontrivial since $n_p>1$. Its <kernel of a group homomorphism> is a <normal subgroup>, so the defining property of a <simple group> makes the <kernel of a group homomorphism> trivial: $\rho$ is an embedding. Now compose with the <sign of a permutation> $S_{n_p}\to\{\pm1\}$. A nontrivial composite would again be <injective> because $G$ is a <simple group>, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho(G)$ lies in the <alternating group> $A_{n_p}$. By <Lagrange's theorem>,
$$
\boxed{|G|\mid |A_{n_p}|=\frac{n_p!}{2}}.
$$
The argument only invokes the factorial formula once $n_p\ge2$; the impossible case $n_p=2$ itself is also excluded by the embedding.
Back to article page
