Group is Subgroup of Itself

Theorem
For any group $$\left({G, \circ}\right)$$, $$\left({G, \circ}\right) \le \left({G, \circ}\right)$$.

That is, a group is always a subgroup of itself.

Proof

 * For all sets $$G$$, $$G \subseteq G$$, that is, $$G$$ is a subset of itself.


 * Thus $$\left({G, \circ}\right)$$ is a group which is a subset of $$\left({G, \circ}\right)$$, and therefore a subgroup of $$\left({G, \circ}\right)$$.