Subgroup is Normal iff it contains Product of Inverses

Theorem
A subgroup $H$ of a group $G$ is normal :


 * $\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$

Necessary Condition
Suppose $H$ is normal.

Let $a b \in H$.

As $H$ is a group, then $x \in H \iff x^{-1} \in H$.

So $b a \in H \iff \left({b a}\right)^{-1} \in H \iff a^{-1} b^{-1} \in H$.

Thus $\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$.

Sufficient Condition
Suppose $a b \in H \implies a^{-1} b^{-1} \in H$.

Then (from the above) we have $\forall a, b \in H: a b \in H \implies b a \in H$.

Let $g \in G$, and let $h \in H$. Let $h = x g$ where $x \in G$.

Now $g h g^{-1} = g x g g^{-1} = g x$.

By hypothesis, $x g \in H \implies g x \in H$.

The result follows.