Is my textbook wrong about this corollary of Sylow's theorem?
Let $G$ be a finite group and $p$ a prime that divides $|G|$. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then $n_p = 1$ if and only if the Sylow $p$-subgroup is normal. Hence, if the number of Sylow $p$-subgroups is one, then $G$ is not simple.
Well, this clearly isn't true if $|G| = p$.