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

 * As $$H$$ is a subgroup of $$\left({G, \circ}\right)$$, $$H$$ is closed under $$\circ$$.

Therefore, $$\forall \left({h_1, h_2}\right) \in H \times H: h_1 \circ h_2 \in H$$.

Therefore, $$H \circ H \subseteq H$$.


 * Let $$e$$ be the identity of $$G$$, and by Identity of Subgroup, of $$H$$.

Then:


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

Hence the result from the definition of set equality.