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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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    $\begingroup$ The condition should be "if the number of Sylow $p$-subgroups is one and $G$ is not a $p$-group." $\endgroup$ – Qiaochu Yuan May 4 '12 at 21:21
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    $\begingroup$ Or, more specifically than Qiaochu's addendum, "and $G$ is not of order $p$" $\endgroup$ – Arturo Magidin May 4 '12 at 21:25
  • $\begingroup$ Alternatively, "then $G$ is not nonabelian simple". $\endgroup$ – Chris Eagle May 4 '12 at 21:28
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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 $p$).

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