Sylow p-Subgroup is Unique iff Normal
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Theorem
A group $G$ has exactly one Sylow $p$-subgroup $P$ if and only if $P$ is normal.
Proof
If $G$ has precisely one Sylow $p$-subgroup, it must be normal from Unique Subgroup of a Given Order is Normal.
Suppose a Sylow $p$-subgroup $P$ is normal.
Then it equals its conjugates.
Thus, by the Third Sylow Theorem, there can be only one such Sylow $p$-subgroup.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $11$: The Sylow Theorems: Example $11.12$: Remark