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 $G/H$.
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 $H \rightarrow G$.
A normal subgroup is a simple and unique way to characterize any homomorphism
When the word "normal subgroup" is mentioned, your first reflex has to be to ask yourself "what's the associated homomorphism".
First to be fully pedantic, let's start with: why are mathematicians so obsessed with homomorphisms?
An isomorphism is a bijective function between two groups (of the same size since it's a bijection) and means that they are the exact same as far as the group structure is concerned. Very strong, and very boring.
An homomorphism however does not have to be a bijection: it can take a larger group and transform it into a smaller image group. Notably, several distinct inputs can map to the same output (non-injective).
The tradeoff is that this smaller group (the image of the homomorphism, which as shown later is isomorphic the quotient G/N) contains a "coarser" group structure than the original group, as it ignores some finer part of the original group (preview: that finer part is the normal subgroup structure). This image structure is simpler because the homomorphism can map multiple input elements to a single output element.
Mathematicians like that because breaking up bigger things into smaller things often allows to tackle the smaller parts in isolation, which often leads to simpler proofs and greater insight. It is a bit analogous to how larger integers can be broken up in to a product of their prime factors (just way more complicated largely because group multiplication is non Abelian).
More ideas on why homomorhpisms are interesting:
- Why the term and the concept of quotient group?
- they show up in Lie Groups. This wiki page actually mentions that some of the isomorphism theorems were actually first formulated by Noether, which makes sense since Lie groups are key to Noether's theorem. The definition of group representations involves an homomorphism.
- the unsolvability of the quintic equation is linked to the simplicity of $A_5$. As mentioned below, "simple" means "without a normal subgroup", or more intuitively "without non-trivial homomorphisms"
Now that we know why homomorphisms are interesting, let's talk about how they relate to normal subgroups
An homomorphism is a function from $G \rightarrow H$, and normally our intuition is that "there is a large number of possible such functions", because there are $order(G) \times orger(H)$ possible arbitrary functions from G to H.
However, in order to keep group structure, this is far from true, and we are much more restricted on our possible choices.
Actually, all we need to fully uniquely specify any homomorphism, is to specify its associated normal subgroup because as shown by the "Fundamental theorem on homomorphisms" mentioned in the section below:
- for any homomorphism, there is one normal group
- for any normal group, there is one homomorphism
This therefore gives a very concrete and natural way of precisely describing the homomorphism in terms of things we understand well: a subgroup of the domain.
Conversely, it also provides clear idea of what a normal group is, because homomorphisms are easy to understand (a function that respects the group operation, that's it!), and now we can understand normal groups in terms of homomorphisms.
So, how do normal groups relate to homomorphisms more precisely?
The precise way in which normal subgroups are related to homomorphisms is given in the aptly named fundamental theorem on homomorphisms. Perhaps the presentation given in the isomorphism theorem page being more understandable.
Here is a commented version of it.
Fundamental theorem on homomorphisms: Let G and H be groups, and let $f: G \rightarrow H$ be a homomorphism. Then:
The kernel of f (noted ker(f)) is a normal subgroup of G
Therefore, each homomorphism uniquely specifies a normal group (the kernel of f).
The image of f is a subgroup of H
This statement is boring. All we usually care about is the image of f, so we might as well always work with the image of f rather than this possibly larger H which contains items we know nothing about given this theorems hypothesis.
The image of f is isomorphic to the quotient group G/ker(f)
This is kind of the converse of the first statement, and says that for every normal group (ker(f)), there is a single homomorphism (the image is isomorphic to G/ker(f)).
Remember that the quotient group is defined as the equivalence classes of:
- the normal subgroup is a class
- the cosets of the normal subgroup are the other classes
Therefore, what this part of the theorem says is that the smaller and simpler output group of the homomorphism ("the image of f") is isomorphic to the above equivalence classes.
More concretely, given a normal subgroup N, we can explicitly construct the corresponding homomorphism as:
$$f(g) = Ng$$
Here is a simplified diagram which illustrates the theorem, which tells us that every homomorphism looks like this:
where:
f is an homomorphism from G to H
N is a normal subgroup of G
e is the identity element of G
$e_H$ is the identity element of H
h1 and h2 arbitrarily selected non-identity elements of H
g1 and g2 are any arbitrarily selected elements such that $f(g1) = h1$ and $f(g2) = h2$.
Since an homomorphism is not necessarily bijective, there are in general several possible choices for $g1$ and $g2$ with that property.
f maps N to $e_H$, $N \cdot g1$ to h1 and $N \cdot g2$ to g2
$N \cdot g1$ and $N \cdot g2$ are two cosets of N when it is multiplied by g1 and g2 respectively.
$N \cdot g1$ contains all elements $g$ such that $f(g) = h1$.
each coset corresponds to one of the elements of H which they map to: N
we see that N is the kernel of f by the definition of kernel, because N is the inverse image of $e_H$
from this it is clear how the structure of the quotient G/N is simpler than the original G: we collapsed the structure of the entire normal group N to a single point! Therefore, an homomorphism is basically a simplification function that ignores the structure of the normal group while doing its transformation
Why the $gN = Ng$ definition of a normal subgroup?
We have to think why this is a necessary and sufficient condition the relation between normal subgroups and homomorphisms to hold.
From the above discussion, we see that if there is an homomorphism, then N maps to the identity of the image ($e_H$).
The necessary side is therefore easy: if we have an homomorphism, because the identity commutes with anything:
$$ f(gN) = f(g)f(N) = f(g)e_H \\ f(Ng) = f(N)f(g) = e_Hf(g) $$
Therefore, suppose that we take another coset like $G1 = N \cdot g1$, which maps to another element of H (h1).
Now for the sufficient, suppose $gN = Ng$. Does that imply that $f(x) = xN$ is an homomorphism? See e.g.: Why do we define quotient groups for normal subgroups only?
Simple group: it looks like a prime number
Now that we know all of this, it becomes clear why simple groups (a group with no Normal subgroups) are analogous to integer primes.
A group without non-trivial normal subgroups (the group itself and the identity) there is no proper homomorphism, i.e., there is no homomorphism except the trivial isomorphism and homomorphism that maps everything to the identity.
And as previously mentioned, an homomorphism breaks up the larger group into two smaller groups (N and G/N) each with part of the original structure.
Therefore simple groups are groups whose structure cannot be broken up in this way: we just can't "factor them out" with an homomorphism.
This is why so much effort was put into the classification of simple finite groups, which turned out to be such an epic result.
Quotient group: it looks like the result of a division
From the above it is also clear why the quotient group is called the "quotient group": it is because it looks similar to dividing an integer G by a factor N.
This is because much like in integer division, we produce a smaller group G/N by taking a larger group G and "dividing" it by a smaller group N.
See also: Why the term and the concept of quotient group?
Group extension problem: what about multiplication?
It is important to note however that this intuition that an homomorphism looks like division only works in one way: we don't really have a good multiplication analogue.
Or more precisely, we don't have a simple algorithm to solve:
Given a finite group N and a simple group S, find all groups $G$ such that N is a normal-subgroup of $G$ and G/N = S.
This happens because groups multiplication is more complex than integer multiplication (notably, non-abelian), so two groups can be composed in more complex ways than two integers, i.e. there is in general more than one possible G that solves the above for some S and N. TODO example of such a case.
You might be tempted to take the direct product of groups as a definition of multiplication, but that alone is not very satisfactory, because as mentioned at When is a group isomorphic to the product of normal subgroup and quotient group?, you would be missing out many corresponding non-trivial "divisions" (homomorphism/quotient).
A slightly better choice would actually be a semidirect product, because the direct product generates a larger group of which both smaller groups are necessarily normal (because of the trivial projection homomorphism), and the semidirect product only needs one of them to be normal. But it is still not general enough.
If we were able to do group extensions algorithmically, then we would able to classify all finite groups, because we have already classified the simple ones.
See also: How is a group made up of simple groups?