# Is my textbook wrong about this corollary of Sylow's theorem?

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 $|G| = p$.

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

## 1 Answer

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 $p$).