Intuition behind normal subgroups

I’ve studied quite a bit of group theory recently, but I’m still not able to grok why normal subgroups are so important, to the extent that theorems like $(G/H)/(K/H)\approx G/K$ don’t hold unless $K$ is normal, or that short exact sequences $1\to N \stackrel{f}{\to}G\stackrel{g}{\to}H\to1$ only holds when $N$ is normal.

Is there a fundamental feature of the structure of normal subgroups that makes things that only apply to normal subgroups crop up so profusely in group theory?

I’m looking here for something a bit more than “$gN=Ng$, so it acts nicely”.

Solutions Collecting From Web of "Intuition behind normal subgroups"

For any subgroup $H$ of $G$, you can always define an equivalence relation on $G$ given by
$$
g_1 \equiv g_2 \iff g_1g_2^{-1} \in H
$$
This lets you define a quotient of $G$ by $H$ by looking at equivalence classes. This works perfectly well, and gives you a set of cosets, which we denote
$$
G/H = \{[g] = gH \mid g \in G\}
$$
However, note that while we started talking about groups, we have now ended up with a set, which has less structure! (There is still some extra structure, e.g. the action of $G$ on the quotient)

We would like to define a natural group structure on this quotient, simply so that we don’t end up in a completely different category. How should this new group structure behave? Well, it seems natural to ask that
$$
[g * h] = [g] *_{new} [h]
$$
so that the map $G \to G/H$ would be a homomorphism (this is, in this context, what I mean by “natural”). So what would this mean? Let’s write it out:
$$
(gh)H = [g * h] = [g]*_{new}[h] = (gH)(hH)
$$
If you work out what these sets are, then you can see that this equation can only be true if we have that $hH = Hh$ for every $h \in G$. But this is exactly the condition that $H$ is normal.

The short answer: $H$ being normal is exactly the condition that we require so that we can put a compatible group structure on the quotient set $G/H$.

Just to expand slightly on Simon Rose’s comment

$H$ being normal is exactly the condition that we require so that we can put a compatible group structure on the quotient set $G/H$.

Suppose for each $x, y \in G$ there is $g \in G$ such that $(x H) ( y H) = g H$, that is, the product of any two left cosets of $H$ is also a left coset.

Take $x = y^{-1}$, so that $1 = y^{-1} 1 y H \in (y^{-1} H) (y H) = g H$, and thus $g H = H$. Thus for every $h \in H$ and $y \in G$ we have
$$
y^{-1} h y = y^{-1} h y 1 \in (y^{-1} H) ( y H) = H,
$$
that is, $H$ is normal.

The normal subgroups of $G$ are all the sets, which appear as kernel of group-homomorphisms $G \rightarrow H$.

Subgroups are the sets, which appear as images of group-homomorphism $H \rightarrow G$.