## Cosets and quotients

I know I’ve been doing a lot of theory here for a while, but just hold on a little longer, it’s about to pay off.

Now we know what a group is and what a subgroup is. Today I want to talk about the cosets of a subgroup in a group .

A subgroup is sort of like a slice through a group. The cosets of a subgroup are like slices “parallel” to the group. We know that passes through the identity of already, so we want to find a “parallel copy” through any other element we choose.

Now, cosets come in two flavors: left and right. We get a left coset by composing every element of on the left with . On the other hand, we get a right coset by composing every element on the right.

Let’s let be the group , and be the subgroup , where is the identity element. For each element of , let’s consider its left and right cosets.

Notice that the element is in both and , since the identity is in . Also notice that if is in then is the same as , and similarly for right cosets. In fact, you should be able to verify that

- is in
- If is in , then is in
- If is in and is in , then is in

and three other similar statements for right cosets. Together, these say that we can separate the elements of into “equivalence classes”. If and are not in the same class then and share no elements at all. But if and are in the same class, and are the same set — the equivalence class itself.

Okay, so we’ve sliced up the group into (left) cosets of the subgroup . If we consider two of them, and , we can multiply everything in the first by to get things in the second, and everything in the second by to get things in the first. These two transformations undo each other, so each coset of is “the same size”. If is a finite group this means that each coset has the same number of elements — the number of elements in . We’ve sliced into a bunch of pieces that never overlap, and all having the same number of elements. This merits some emphasis.

If is a finite group and is a subgroup, then the number of elements in must divide the number of elements of !

Now we’ve got a set of cosets of , which we write . What’s really nice is that sometimes *this set is a group too*! The natural idea to multiply two cosets is to take an element of each and multiply *them* and see what its coset is. Unfortunately this doesn’t always work. The answer might depend on which elements we choose

To see what goes wrong, let’s pick two elements in the first coset. We know that is in , and we can see that so is if is an element of . We just pick again from . Now we multiply to get for one choice and for the other. These are only in the same coset if there is some in so that . That is, so that . The requirement is this: for every in and for every in the composition must land back in . We call a subgroup with this property “normal”.

So if is normal, then our naïve idea for how to multiply two cosets does work right, and doesn’t depend on how we choose the element of each coset to multiply: . You should verify that this composition on the set of cosets of in actually satisfies the group axioms. We call this the “quotient group” of by , or “ modulo “.

Other exercises:

- Check to see that in the example given above (where is ) that the subgroup isn’t normal. Find one that is, and see what the quotient is.
- Show that any subgroup of an abelian group is normal.
- Consider the subgroup of (addition as the composition) consisting of all multiples of 12. Call it . What is ? What if we change 12 to any other number ?

Please dont take lack of comments as indication of lack of interest. Keep it coming! Your efforts are appreciated.

Comment by Astro | February 17, 2007 |

Could you go into more detail about why that last part is true? (By the way, there’s a typo in the sentence I bolded.)

Comment by Jon | December 20, 2007 |

Well, I’ve fixed that typo. A lot of things in old posts seem to break when I’m not looking…

Basically it’s just the definition of being in the same coset. Elements and are in the same coset of if and only if there is an with .

Comment by John Armstrong | December 20, 2007 |

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