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 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.