Group is Subgroup of Itself

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

Then:
 * $\left({G, \circ}\right) \le \left({G, \circ}\right)$

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

Proof
By Set is Subset of Itself, we have that:


 * $G \subseteq G$

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