Why do normal subgroups "work"?

Question

This is something I've wondered about for a while now; What is so special about normal subgroups that makes modding out by them "act nice"?

I understand the proofs for things like the first isomorphism theorem, but all these proofs really seem to do is verify that we can mod out by any sub-group satisfying $gNg^{-1} = N$ and end up with another group. I get the verification; that's not what I'm asking about.

It's possible to mod out by any subgroup, but what's so special about the property $gNg^{-1} = N$ that makes it so that we can "move" $N$ around inside $G$ in a way that satisfies the group axioms? Is there something deeper about the normality property that makes this make sense?

Answer 1

Concerning your comment about "the group consisting of ways to move N inside of G", the group structure is actually being given to a way of moving the cosets of $N$ around rather than $N$ itself.

A standard way of moving subgroups themselves around is by conjugation:- $H$ going to $g^{-1}Hg$.

Here one can see something very special about normal subgroups. They are precisely the subgroups which do not move around at all!

Furthermore this is the property used in standard proofs that the group operation on cosets is well defined.

Answer 2

A general quotient set is done by an equivalence relation $\sim$ on a set $S$, and we can informally think about $S/\sim$ as the set obtained by keeping the elements but replacing the original equation of $S$ by $\sim$, so that $x\sim y$ in $S$ means $x=y$ in $S/\sim$.

Now, indeed any subgroup $H\le G$ determines an equivalence relation by taking the partition of left (or right) cosets, so nothing stops us to define the quotient set $G/H$ as $G/\sim_H$ where $x\sim_H y\iff y^{-1}x\in H$.

However, in order for it to inherit the group structure, we also need the induced equivalence relation to respect the group operation, and in particular we need $gg^{-1}=1$ to hold.
But in $G/H$ we will have $h=1$ for any $h\in H$, so we must have $ghg^{-1}=g1g^{-1}=gg^{-1}=1$ in $G/H$, which really means $ghg^{-1}\sim_H 1$ in $G$, i.e. $ghg^{-1}\in H$.

It turns out that this additional condition (of being normal) on $H$ is sufficient as well: in that case the group structure is inherited to the quotient set.

A more formal way is that normal subgroups are exactly the kernels of some homomorphism (where kernel is the preimage of $1$).