# Why normality is important?

1. Sep 21, 2011

Why normal subgroups are very important in group theory? what's so good about them? I know that normality plays an important role in defining a simple operation on co-sets of a group and it also plays an important role in defining the quotient group, but what I'm curious to know is the history behind normality and the applications of normal subgroups and its consequences in modern Algebra.

2. Sep 21, 2011

### micromass

In my opinion, normal subgroups are only important because you can quotient them out. That is: you can give sense to the quotient group. So the question becomes: why is the quotient group important?

Well, often a group contains too much information. And a quotient group can reduce this information by a lot. I believe that group theory can be done without quotient groups, but it'll be a lot uglier!

From the point-of-view of universal algebra, the normal subgroups correspond to the congruence relations.

3. Sep 22, 2011

well, let me see if I have understood the concept correctly. suppose we are in the group integers under addition. since any subgroup of Z is of the form nZ and every subgroup of it is normal, since Z is Abelian, the quotient group is defined for any subgroup of Z, like nZ, and contains the elements of Z that are the same in view point of being dividable by n.

so, we define an equivalence relation on G that has come from our experience with integers under addition and then we define the co-sets of a subgroup of G as being equal in the sense that they have a specific property that makes them equivalent(congruent).
Herstein beautifully describes that the relation a~b iff ab-1 is in H (where H is a subgroup of G) gives us an equivalence relation on G that is similar to the congruence relation on integers and then the co-sets aH or Ha are in fact like [a] in set theory, and G/H is the same as the set G/H={aH:a in G} (aH=Ha iff H is normal). then if H was normal, we could define a naive operation on it by defining aH.bH=(a.b)H and this operation is well-defined and turns the set G/H into a new group called the quotient group. Am I right?

Can we define another equivalence relation on G and do the same process on it? I mean there can be thousands or millions of equivalence relations on G. Why we are interested to only this particular equivalence relation? Or Are there other similar concepts in group theory that will come later?

Last edited: Sep 22, 2011
4. Sep 22, 2011

### micromass

Why are we only interested in this particular equivalence relation? Because it is the only congruence relation. That is: it is the only relation such that the quotient set is a group.

Let $\sim$ be a congruence relation. We say that it is a congruence if
• $a\sim b,~c\sim d~~\Rightarrow~~ ac\sim bd$
• $a\sim b~~\Rightarrow~~a^{-1}\sim b^{-1}$

Thus something is a congruence relation if it is compatible with the group operation.
It turns out (and it is not so hard to prove) that the congruence relations are exactly of the form $ab^{-1}\in N$ for N normal.
In rings, the congruence relations are exactly of the form $a-b\in I$ with I an ideal.

In other structures (such as lattices, semigroups, etc.) there is no easy characterization of congruence relations.

But really, it is not the normal subgroup that interests us, but rather the congruence relation.

5. Sep 22, 2011

Thanks for the detailed answer. It was very educative for me.

How can you prove that $ab^{-1}\in N$ is the only congruence relation that is compatible with the group operation?

6. Sep 22, 2011

### micromass

A terminology fix: a congruence relation is already compatible with the group operation. That is: a congruence relation is by definition an equivalence relation that is compatible with the operation.

The trick is to set N the equivalence class of e. That is, set

$N=\{x\in G~\vert~x\sim e\}$.

It is easily proven that N is a subgroup. To prove that N is normal, take x in N. Then $gxg^{-1}\sim geg^{-1}=e$. Hence $gxg^{-1}\in N$.

We need to prove that $a\sim b$ if and only if $ab^{-1}\in N$.
Indeed: take $a\sim b$, then $ab^{-1}\sim bb^{-1}=e$, such that $ab^{-1}\in N$.
On the other hand, take $ab^{-1}\in N$, then $ab^{-1}\sim e$. Hence $ab^{-1}b\sim eb$. Thus $a\sim b$.

7. Sep 22, 2011

### HallsofIvy

Historically, group theory had its origins in Galois's proof that there exist polynomial equations having roots that cannot be expressed in terms of nth roots. The crucial point is that a polynomial equation is "solvable by radicals" if and only if its Galois group is "solvable" which is defined as:
There exist a sequence of groups $e\subset G_1\subset G_2\subset \cdot\cdot\cdot \subset G_{n-1}\subset G$ such that each $G_i$ is a normal subgroup of $G_{i+1}$ and the quotient group $G_{i+1}/G_i$ is commutative.

8. Sep 22, 2011

### mathwonk

the kernel of a homomorphism is normal. hence if G has no normal subgroups, then every homomorphism from G is injective. this is a very useful fact.

i.e. normal subgroups are important because homomorphisms are important.

9. Sep 25, 2011

### Bacle

To add something that may be relevant here, while, As Mathwonk mentioned, given

a homomorphism h: G-->G' , then the kernel (not Sanders, but the preimage of the identity)

is normal in G, but, given any normal subgroup N of G , there is a homomorphism from

G into some other group, of which N is the kernel, i.e., the "natural" one q: G-->G/N.

10. Oct 3, 2011

### Deveno

normal groups are important because they are kernels.

that is, every kernel of a homomorphism φ:G-->G' is a normal subgroup, and conversely, every normal subgroup N of G is the kernel of the homomorphism G-->G/N which sends g-->gN.

this tells us the the homomorphic images of a group G are just factor groups G/N for some N normal in G, "in disguise".

you can think of it this way: suppose we have a group G and we want to ignore what some of the elements of G do. well, we don't want to "mess up" the group operation, and the only way this doeesn't happen, is if the elements we want to ignore form a normal subgroup.

think of Dn, the symmetry group of an n-gon in R^2. if all we are interested in, is whether or not we have a "right-hand n-gon" (top side up), or a "left-hand n-gon" (top side down), then we don't care about how "rotated we are" (rotations don't affect the orientation). so we'd like to think of rotations as being "normal", that is, they preserve the orientation. so we'd like to "collapse" all rotations to the identity map.

and the rotation group is normal, so we can do this. we can "take out the r's" in the product:

(r^j)(s^k)*(r^m)(s^n), and just compute (s^k)(s^n), without fear of inconsistency (what we are really doing is sending (r^j)(s^k) to the coset <r>s^k).

but the reflections are not so well-behaved: two reflections composed make a rotation, but not necessarily the rotation you would hope:

(r^k)s*(r^m)s is not necessarily r^(k+m), so we can't just "take out the s's".

(try this with a hexagon, for example).

normal groups play the role of "factors" for groups. simple groups (groups with no normal subgroups) are analogous to "prime numbers".

11. Oct 3, 2011

### lavinia

Every subgroup determines a coset space ( left or right) . If the subgroup is normal then the coset space is naturally a group. Conversely if the coset space is a group then the subgroup is normal. This is the key.

A group can be thought of a being built up from a normal subgroup and its quotient group. This helps one to understand the structure of the group. One part of this analysis is to look at the way in which lifts of the elements of the quotient group act on the normal subgroup under conjugation. For instance the dihedral group of order 8 has a cyclic normal subgroup of order 4 and the quotient is Z2. A lift of the non-zero element of Z2 to D8 acts by conjugation on Z4 by a -> a^3.

Last edited: Oct 3, 2011