Unique Subgroup of a Given Order is Normal

Theorem
Let a group $G$ have only one subgroup of a given order.

Then that subgroup is normal.

Proof
Let $H \le G$, where $\le$ denotes that $H$ is a subgroup of $G$.

Let $H$ be the only subgroup of $G$ whose order is $\left|{H}\right|$.

Let $g \in G$.

From Conjugate of Subgroup is Subgroup:
 * $g H g^{-1} \le G$

From Order of Conjugate of Subgroup:
 * $\left|{g H g^{-1}}\right| = \left|{H}\right|$

But $H$ is the only subgroup of $G$ of order $\left|{H}\right|$.

Hence any subgroup whose order is $\left|{H}\right|$ must in fact be $H$.

That is, $g H g^{-1} = H$.

The result follows from Subgroup equals Conjugate iff Normal.