Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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".

share|improve this question
5  
Otherwise, the quotient is not a group, so most of the interesting questions do not even make sense to ask in that case. –  Tobias Kildetoft Apr 30 at 18:54
    
$(G/H)/(K/H)\cong G/K$ and $N\to G\to H$ being short exact are impossible without normality. –  blue Apr 30 at 18:54
    
@TobiasKildetoft Are there any other reasons or is that pretty much it? –  Alyosha Apr 30 at 19:00
1  
Since normal subgroups are precisely the ones that allow a natural group structure on the quotient, in some sense, it is $the$ reason. –  Tobias Kildetoft Apr 30 at 19:01
2  
A version of this question that I liked when I first learned abstract algebra: Why isn't a circle a ring? Why isn't $\mathbb{R}/\mathbb{Z}$ a ring? –  Jack Schmidt Apr 30 at 19:03

3 Answers 3

up vote 8 down vote accepted

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 $G/H$.

share|improve this answer
    
In my opinion, this hits the nail exactly on the round flat part. –  MJD Apr 30 at 19:30
    
I prefer to think about this in the notation of modular arithmetic. We can always define an equivalence relation $\equiv$ meaning "are in the same coset" given any subgroup, and this equivalence relation is compatible with the group operation iff that subgroup is normal. –  Jack M Apr 30 at 20:45

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 $H \rightarrow G$.

share|improve this answer
    
IMHO this is the correct answer. –  Steven Gubkin Apr 30 at 20:45

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.