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.