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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 $|G| = p$.

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    Alternatively, "then $G$ is not nonabelian simple".2012-05-04

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Yes, there is the hypothesis that $|G|$ is not a prime that needs to be added; other than that, it is of course correct (if $|G|=p^k$ with $k\gt 1$, then we know the group is not simple as it always has a normal subgroup of order $p$).