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.