Group is Normal in Itself

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

Then $\left({G, \circ}\right)$ is a normal subgroup of itself.

Proof
First we note that $\left({G, \circ}\right)$ is a subgroup of itself.

To show $\left({G, \circ}\right)$ is normal in $G$:
 * $\forall a, g \in G: a g a^{-1} \in G$

as $G$ is closed by definition.

Hence the result.