# Group is Subgroup of Itself

## Theorem

Let $\struct {G, \circ}$ be a group.

Then:

$\struct {G, \circ} \le \struct {G, \circ}$

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

## Proof

By Set is Subset of Itself, we have that:

$G \subseteq G$

Thus $\struct {G, \circ}$ is a group which is a subset of $\struct {G, \circ}$, and therefore a subgroup of $\struct {G, \circ}$.

$\blacksquare$