Unique Subgroup of a Given Order is Normal

Theorem
If a group $G$ has only one subgroup of a given order, then that subgroup is normal.

Proof
From Conjugate of Subgroup is Subgroup, we have that if $H \leq G$ then $ g H g^{-1} \leq G$.

From Order of Conjugate of Subgroup, the conjugate of a subgroup has the same order as the subgroup.

Let $H \le G$ such that $H$ is the only subgroup whose order is $\left|{H}\right|$.

Let $g \in G$.

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

But any subgroup whose order is $\left|{H}\right|$ must in fact be $H$, because $H$ is the only subgroup of $G$ of that order.

From Normal Subgroup Equivalent Definitions: 5, it follows that $H$ is normal.