Hi all! Today we’ll be talking about quotient groups.

We can ask: if is a homomorphism, is isomorphic to a subgroup of ? Certainly it can’t be the whole group itself, because the kernel of could be nontrivial, i.e., more than one element could map to the identity. (Unless the homomorphism is injective, because then is trivial.) It turns out that the subgroup of which is isomorphic to is something called the *quotient group* of .

Another, perhaps more simpler way of motivating quotient groups is as follows. Suppose is a group, and . Can we find a group and a homomorphism such that is isomorphic to ? (Because is normal, you obviously need to be a normal subgroup of .)

We can choose the *underlying set* of to be the set of all left cosets of in . The map takes . But can become a group (i.e. what’s the multiplication)? Suppose . Then define . Except this isn’t well-defined!

The reason is, if and , then and . In order for this multiplication to be well-defined, we need if and .

So we need . Well, and can be written as and for , because, well, ; since , it’s also in , and similarly for . Now, . We need this to be of the form for some . We want to shift the to before the .

What if is normal? Well, then . Suppose we write , so that . Then (by left multiplication), so let’s replace this equality above, to get . Since , and , we see that as well, and thus, .

So in order for the multiplication to be well defined we need the subgroup of to be normal. Now, because this makes the multiplication well-defined, we can use the fact that is a group, and state:

**Lemma: ***If is the set of all left cosets of , the multiplication gives a group structure on if is normal.*

**Definition: **The set above is called the *quotient group* of modulo , and is denoted .

**Proposition: ***The map defined by is a surjective group homomorphism. The kernel of is .*

*Proof: *Obviously is surjective (because if , then for some , which means that it can be pulled back to some element of ). It’s a group homomorphism (this is by definition!). Now we want to show that .

Suppose . Then , because is closed under multiplication. Now, is the identity in (because if , then ). So any is mapped to the identity of , or, in other words, . What this shows is that .

Suppose . Then , because, as we said above, is the identity in . So , and therefore, (because if not then they can’t be equal!). So . Combining these two shows that . QED.

In other words actually satisfies what we talked about in the introduction to the talk. Actually, what this shows is that *every normal subgroup is the kernel of some group homomorphism*.

It’s finally time to explain the notation . Recall that is an abelian group, so every subgroup is normal. Since is a normal subgroup of , we can consider . We’ll see that this is exactly the cyclic group of order . The elements of are cosets for , and *must* be less than , because otherwise this’ll be the same as . So there are elements of , and the addition is modulo . And, well, this is *exactly* the cyclic group of order !

Here are trivial examples: has one element, namely the identity. Also, is just itself, because if , then , so .

Here’s an interesting example. Suppose , and let . What is ? Suppose is the group of complex numbers, and suppose is the unit circle, i.e., the collection of all complex numbers of the form , for . So, define as . Well, if , then , and must be an integer. So the kernel of consists of the integers, i.e., . Then the map defined by is an isomorphism. The reason will be revealed in the next post, which will be on the isomorphism theorems (for groups)!

Have fun,

SKD