Review: Normal Subgroups

$\textbf{Definition}$: A subgroup $N \leq G$ is normal if it is invariant under conjugation. i.e. $$\forall x\in{N}, \quad \forall g\in{G}, \quad gxg^{-1}\in{N}$$ [Remember: invariance is an important concept ubiquitous in mathematics.]
The following are some of the conditions equivalent to $N \unlhd G$:

(i) $ \forall g\in{G}, \quad gNg^{-1}= \{gxg^{-1}|g\in{G},x\in{N} \} \subseteq N$
(ii) $ \forall g\in{G}, \quad gNg^{-1}= N$
(iii) $ \forall g\in{G}, gN=Ng$ (left and right cosets are equal)
(iv) $N_G(N)=\{g\in{G}|gNg^{-1}=N\}=G$
(v) the operation of left (right) cosets of $N$ in $G$ makes the set of left (right) cosets into a group
(vi) $N$ is a union of conjugacy classes of $G$
(vii) $\exists \varphi:G \rightarrow H$, a homomorphism for which $N=ker\varphi$

Item (vii) demonstrates some of the importance of normal subgroups; they are a way to internally classify all homomorphisms defined on a group. For example, a non-trivial group is $\textit{simple}$ iff it is isomorphic to all of its non-trivial homomorphic images. This can be seen by means of the following:
Let $\varphi : G \rightarrow H$. By the first isomorphism theorem, $$im\varphi \cong G/ker\varphi $$ If $G$ is simple, then $ker \varphi$ (which is normal) is either $G$ or $\{e\}$. Thus, $G$ will be isomorphic to all of its non-trivial homomorphic images.

Item (vi) is justified because conjugation defines an equivalence relation; so conjugacy classes partition $G$. (This appears later when deriving the class equation). Normal subgroups are invariant under conjugation, so they can be viewed as a union of conjuacy classes of $G$.
In practice, it seems to be most useful to show item (i) or (ii). Pick $x\in{N}$, then show that for arbitraty $g\in{G}$, $gxg^{-1}\in{N}$. We can think of normality as a sort of partial commutativity that says: upon conjugating an element, the possible resulting elements will be bounded within the subgroup $N$. For readers familiar with linear algebra, notice that in the group of linear transformations $GL_n(F)$, conjugation is the same as a change of basis $A \mapsto QAQ^{-1}$.