Trivial Subgroup is Subgroup

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

Then the trivial subgroup $\left({\left\{{e}\right\}, \circ}\right)$ is indeed a subgroup of $\left({G, \circ}\right)$.

Proof
Using the One-Step Subgroup Test:


 * $(1): \quad e \in \left\{{e}\right\} \implies \left\{{e}\right\} \ne \varnothing$
 * $(2): \quad e \in \left\{{e}\right\} \implies e \circ e^{-1} = e \in \left\{{e}\right\}$