Subgroup of Order 1 is Trivial

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

Then $\left({G, \circ}\right)$ has exactly $1$ subgroup of order $1$: the trivial subgroup.

Proof
From Trivial Subgroup is Subgroup, $\left({\left\{{e}\right\}, \circ}\right)$ is a subgroup of $\left({G, \circ}\right)$.

Suppose $\left({\left\{{g}\right\}, \circ}\right)$ is a subgroup of $\left({G, \circ}\right)$.

From Group is not Empty, $e \in \left\{{g}\right\}$.

Thus it follows trivially that $\left({\left\{{g}\right\}, \circ}\right) = \left({\left\{{e}\right\}, \circ}\right)$.

That is, $\left({\left\{{e}\right\}, \circ}\right)$ is the only subgroup of $\left({G, \circ}\right)$ of order $1$.