Group Theory – Intuition Behind Normal Subgroups

group-theoryintuitionnormal-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".

Best Answer

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 $G/H$.