Sylow p-Subgroup is Unique iff Normal

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
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