Sylow p-Subgroup is Unique iff Normal

Theorem
A group $G$ has exactly one Sylow $p$-subgroup $P$ $P$ 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.