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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 $|G| = p$.

  • 4
    The condition should be "if the number of Sylow $p$-subgroups is one and $G$ is not a $p$-group."2012-05-04
  • 3
    Or, more specifically than Qiaochu's addendum, "and $G$ is not of order $p$"2012-05-04
  • 0
    Alternatively, "then $G$ is not nonabelian simple".2012-05-04

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