Trivial Subgroup and Group Itself are Normal

Theorem
For every group $G$, the subgroups $G$ and $\left\{{e}\right\}$ are normal, where $e$ is the identity of $G$.

Proof
First we note that:
 * the trivial group $\left\{{e}\right\}$ is a subgroup of $G$;
 * the group $G$ is a subgroup of itself.

So both $G$ and $\left\{{e}\right\}$ are subgroups of $G$. All we need to do is show they are normal.

So:
 * $\forall a, g \in G: a g a^{-1} \in G$ as $G$ is closed by definition.


 * $\forall a \in G: a e a^{-1} = a a^{-1} = e$.

Hence the result.