Factor Groups

Last time we learned about cosets and at the end I alluded to an extremely important group called a factor group. I mentioned that if H is a subgroup of a group, G, then the set of all left cosets of H in G is a group only when H has some mysterious property. The condition that we require is that H is normal, and this condition of normality is invented precisely to allow this set of cosets to form a group. I not only want to convince you that this factor group exists, but also that this normality condition is required. I'm going to start by defining and discussing normal subgroups.

Definition: Normal Subgroup
Let G be a group and N be a subgroup of G. Then N is a normal subgroup of G if aN = Na ∀a∈G and is written N⊳G.

Basically, a subgroup is normal when left cosets are the same as right cosets. I know that it seems like a strange condition, but it turns out to work nicely. Now we first should note what this is not saying. The condition that aN = Na does not say anything about commutivity. New students often think that this means that for n∈N, a∙n = n∙a or something like that - it says nothing about the commutivity of elements - it only says that the sets are equal. I'm not going to work out an entire example, but I do want to show how the cosets can be equal without commutivity. Consider S₄ and N = {ε,(12)(34),(13)(24),(14)(23)}. I'm going to compute (123)N and N(123). Observe that (123)(12)(34)=(134)=(14)(23)(123), (123)(13)(24)=(243)=(12)(34)(123), and (123)(14)(23)=(142)=(13)(24)(123). Because of this, we then get that (123)N = N(123) however none of the elements of N (except for ε) commute with (123). We haven't yet proven that N is normal - that would require showing that πN = Nπ ∀π∈S₄. However, it does turn out to be true that N is normal in S₄. We are now prepared to make factor groups.

Theorem: Factor Groups form Groups
Let G be a group and let N⊳G. Then the set G/N = {aN : a∈G} is a group under the operation of (aN)∙(bN) = (a∙b)N. We call G/N a factor group.

I'm not going to formally prove it, but its important to talk about why the normality of N is necessary. G/N easily passes closure and associativity without normality - that is that (aN)∙(bN)=(a∙b)N∈G/N and [(aN)∙(bN)]∙(cN) = ((a∙b)∙c)N = (a∙(b∙c))N = (aN)∙[(bN)∙(cN)] ∀aN,bN,cN∈G/N. Also, it is pretty clear that eN (or just N) is the identity, whereas a^-1N is the inverse of aN. However, we defined the operation on G/N by (aN)∙(bN) = (a∙b)N and the problem is that G/N is a set of cosets and we defined the operation between these cosets based on their coset representatives. But, as we already learned, any element of a coset can serve as its coset representative, so we must ensure that the operation on G/N is independent of choice of coset representative - that is that if a₁N = a₂N and b₁N = b₂N, then we should ensure that (a₁∙b₁)N = (a₂∙b₂)N. I am going to write that part out formally in an effort to make it as clear as I can.

Proof:
Choose a₁,a₂,b₁,b₂∈G such that a₁N = a₂N and b₁N = b₂N. Notice that a₁∈a₁N=a₂N and also that b₁∈b₁N=b₂N. The first of these gives us that ∃n∈N such that a₁=a₂∙n, and similarly, the second of these gives us that ∃m∈N such that b₁=b₂∙m. Also, remember that when x∈N, xN = N = Nx. Now we are prepared to show that (a₁∙b₁)N = (a₂∙b₂)N. Observe that (a₁∙b₁)N = (a₂∙n∙b₂∙m)N = (a₂∙n∙b₂)(mN) = (a₂∙n∙b₂)N = (a₂∙n)(b₂N) = (a₂∙n)(Nb₂) = a₂(nN)b₂ = a₂Nb₂ = a₂(Nb₂) = a₂(b₂N) = (a₂∙b₂)N. Finally, this gives us that coset multiplication does not depend on the choice of coset representative.

You'll see that I only barely needed normality in that proof, but twice I used the fact that bN = Nb. And you can try as hard as you like, but there is no way to prove this without the normality of the subgroup. In fact, I'm not going to prove it, but it turns out that if N is a subgroup of G and G/N forms a group under the above-defined operation, then it is guaranteed that N is normal. This idea of a factor group is a bit of an abstract thing and can be hard to understand, but next time I'm going to give an example to help clarify.

References

Previous Related Post: Cosets

Text Reference: Gallain Chapter 9

Wikipedia: Quotient Group

Wolfram Mathworld: Quotient Group

Planet Math: Quotient Group