Product of Subgroup with Itself

Theorem
Let $\left({G, \circ}\right)$ be a group.

Then $\forall H \le \left({G, \circ}\right): H \circ H = H$

Proof

 * From Groupoid Subset Product with Self, we have:
 * $H \circ H \subseteq H$


 * Let $e$ be the identity of $G$

By Identity of Subgroup, it is also the identity of $H$.

So:


 * $h \in H \implies e \circ h \in H \circ H \implies h \in H \circ H \implies H \subseteq H \circ H$.

Hence the result from the definition of set equality.