Subgroup is Normal iff it contains Product of Inverses

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


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

Proof

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


 * 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.