Today I have more of a specific idea of what I want to talk about, since we’ve got some background out of the way. We’re going to define quotient groups and look at some examples. The quotient operation is dual to the subgroup operation: a quotient group is, effectively, what you get when you “trivialize” a subgroup. This general idea adapts itself to many other mathematical structures, like, as we’ll see, topological spaces. For groups, however, it’s only possible to do this for certain kinds of subgroups, and so a little background is needed. Let’s get started!

First observe that the multiplication operation naturally extends to *subsets* of a given group. We define , where and are subsets of a group , to be the set . We abbreviate to . This operation is clearly associative and has an identity . On the other hand, the cardinality of is at least as big as the larger of (prove this), so there aren’t inverses, in general. We call a set with an associative binary operation and an identity a **monoid**.

If is a *subgroup* of , then we call the sets the **left cosets** of . If , then , but if not, the cosets are disjoint; thus, the left cosets partition . The set of equivalence classes (or of “distinct” cosets) is written . Similarly, the **right cosets** of are the sets , and the set of distinct right cosets is .

Only the coset is a subgroup of , because it alone has the identity. But that’s okay. The cosets are supposed to be the effect of taking “cross-sections” of that are “parallel” to . Letting , we get three other left cosets: and . We call the number of distinct cosets (cardinality, if necessary) the **index** of in , and write this as .

**Theorem** (Lagrange). *For any , we have .*

**Proof**. We remarked above that . In this case, it’s easy to see that because left-multiplying by is a bijection between them. Thus, all cosets have cardinality . Since the distinct cosets partition , we have . This holds for infinite groups, too, in the sense of cardinality. It also holds for right cosets, for obvious reasons.

**Corollary**. *The order of a subgroup divides the order of the group.*

And finally, the **order** of an element is the smallest such that , or “infinity” if this never happens. Then since each element generates a cyclic subgroup of the same order as itself:

**Corollary**. *The order of an element divides the order of the group.*

This is really cool because it puts severe restrictions on the kinds of subgroups and elements you’re allowed. We’ll see this combinatorial approach again with the Sylow theorems.

For now, let’s look at quotients. We’d like to make into a group, but multiplication kind of stops working: which is not a left coset, in general. The key realization is that multiplication still works for certain *kinds* of cosets.

We define a **normal** subgroup to be a subgroup such that for all . We call the operation **conjugation** by ; so a normal subgroup is one that’s closed under conjugation. Now, if is normal, then : the right and left cosets canonically coincide. In particular, , so the product of two cosets is again a coset.

Thus, for a normal subgroup, the set has a group structure induced from its monoid structure, with . We have to check this is well-defined, i.e., it doesn’t matter which representative of an equivalence class you pick. Just observe that if , then , and , where . We call with this operation the **quotient group** of by .

Let’s look at again (if anyone has another group to suggest, we can do that, too). is not normal because . But is: any rotation is already in the group, and any flip conjugates itself. There are two cosets, the other one being . Calling the first and the second , we get . This is the cyclic group of order two/symmetric group on two symbols/dihedral group of order two/whatever: up to isomorphism, there’s only one group of order two.

As another example, consider the infinite cyclic group and the subgroup generated by . Since is abelian, conjugation doesn’t do anything, so every subgroup is normal. Thus, we can form the quotient group , which is just the cyclic group of order ! This explains the notation from earlier.

Now let’s look at something cool, called the First Isomorphism Theorem. (There are other Isomorphism Theorems, but this is the cool one). It both characterizes and streamlines quotients of groups, and for the most part allows us to avoid thinking about cosets at all. First, some terminology. Let be a homomorphism. As with all maps, the **image** of , , is the set . We can also define the **kernel** of , , as , the preimage of the identity of .

**Theorem** (First Isomorphism). *Let be a homomorphism. Then its kernel is a normal subgroup of , its image is a subgroup of , and .*

**Proof**. The kernel of contains ; if , then , so ; and if , then , so . So the kernel of is a subgroup of . Check for yourself the similar statements for . If , then , so , and is normal.

The isomorphism we expect is the obvious one. Let . Define by . Is this well-defined? If , then for some . Then since is in the kernel. Thus, the mapping does not depend on the choice of coset in an equivalence class. It is injective: if , then , so . Then , so . Finally, it is surjective: for any , we have for some , so as well. Thus, is an isomorphism.

It’s worth pointing out that the converse basically holds: every normal subgroup can be characterized as the kernel of a homomorphism. Let be normal in , and let be defined by . Then clearly is the kernel of this map.

There are so many examples for this it isn’t even funny. An easy one is the map given by “taking everything mod .” The kernel of this map is the set of multiples of , which is just . So we proved that . I hope you are proud of our marvelous achievement.

Here’s a bigger example. Let be the group of real numbers under addition. Then the integers are a normal subgroup. We can define a map with as the kernel, given by sending each real number to its “fractional part,” everything past the decimal point. Then is isomorphic to the image of this map, which is the “reals mod 1,” and is basically given by putting the interval in a circle so that touches .

Another example is the absolute value function on . This is a surjective homomorphism to . Its kernel is the set of complex numbers with absolute value 1, which is just the circle . So .

Next time, we’ll use these results to talk about free groups and the generators-and-relations presentation of a group. This should give us all the background we need for the Banach-Tarski Paradox!

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