Apologize for the potentially long question. For a long time, I always thought there is some connection between group conjugation and normal subgroup. I have no problem following the formal proofs and solving problems (perhaps). Nevertheless, I would like to have a more "natural" understanding of the concepts, whatever that means.
I used to introduce myself to the concept of normal subgroup by the following question:
We know that given a homomorphism $f:G \to H$, $\ker(f)$ is always a subgroup of $G$. But given an arbitrary subgroup of $G$, say $N$, is it true that $N$ is the kernel of some group homomorphism?
The answer is "No" and counterexamples could be easily constructed using, say, $S_n$. Then I see that if I want this claim to be true, I really need that $gNg^{-1} = N$.
This is all good, until the other day I came across some lecture notes online, saying that one conjugation in group resembles similar matrices. Since we understand similar matrices as "different matrices representing the same thing under different basis", we can use the same idea to understand group conjugation. Indeed, the "conjugate-by-$x$" map is an automorphism of $G$.
I thought this was really interesting, because it essentially is saying that "normal subgroups are the subgroups that are invariant under conjugation". That is, $N$ being normal means that no matter what kind of "transform" ($gNg^{-1}$) it goes through, it stays the same.
Now the final piece that I couldn't get is this
Why "being invariant under conjugation" is critical to "being kernel of some homomorphism"?
I hope this is a meaningful question to ask.
Let $G$ and $H$ be multiplicative groups with multiplicative identity $``e"$. We will let context indicate which group $e$ belongs to. Suppose $f:G \to H$ is a homomorphism and let $K$ be the kernel of $f$; i.e., $K = \{x \in G : f(x) = e\}$. For every $k \in K$ and every $g \in G$, $f(gkg^{-1}) = f(g) f(k) f(g^{-1}) = f(g) e f(g^{-1}) = e$. With just a bit more work, it follows that, for every $g \in G$, $gKg^{-1} = K$.
Suppose $N \subseteq G$ is invariant under conjugation. Then the set of cosets of $N$, $\; G/N, \;$ is a group. In fact, the mapping $f:G \to G/N$, defined by $f(g) = gN$, is a group homomorphism. Clearly $N$ is the kernal of $f$.