Let . Then becomes a group under coset multiplication. Define the quotient map (or canonical projection) by
Lemma. If , the quotient map is a surjective homomorphism with kernel H.
Proof. If , then
Therefore, is a group map.
Obviously, if , then . Hence, is surjective.
Finally, I'll show that . If , then , and H is the identity in . Therefore, , i.e. .
Conversely, suppose . Then , so , so . Therefore, , and hence .
The preceding lemma shows that every normal subgroup is the kernel of a homomorphism: If H is a normal subgroup of G, then , where is the quotient map. On the other hand, the kernel of a homomorphism is a normal subgroup. Hence:
Normality was defined with the idea of imposing a condition on subgroups which would make the set of cosets into a group. Now an apparently independent notion --- that of a homomorphism --- gives rise to the same idea! This strongly suggests that the definition of a normal subgroup was a good one.
You can think of quotient groups in an even more subtle way. The general theme is something like this. In modern mathematics, it is important to study not only objects --- like groups --- but the maps between objects --- in this case, group homomorphisms. The maps, after all, describe the relationships between different objects. (This theme is elaborated in a branch of mathematics called category theory.)
It turns out that more is true. In a sense, the maps carry all of the information about the objects; one could even be perverse and "build up" objects out of maps! I won't go to such extremes, but in some cases, an object can be characterized by certain maps. Here's an important example.
Theorem. ( Universal Property of the Quotient) Let , and let be a group homomorphism such that . Then there is a unique homomorphism such that the following diagram commutes:
(To say that the diagram commutes means that .)
Proof. Define by
This is forced by the requirement that , since plugging into both sides yields , or .
I need to check that this map is well-defined. The point is that a given coset may in general be written as , where . I must verify that the result or is the same regardless of how I write the coset.
(If in this situation, then a single input --- the coset --- produces different outputs, which contradicts what it means to be a function.)
Suppose then that , so for some .
This shows that is indeed well-defined.
I was forced to define as I did in order to make the diagram commute. Hence, is unique.
Now I'll show that is a homomorphism. Let . Then
Therefore, is a homomorphism.
The universal property of the quotient is an important tool in constructing group maps.
The map you construct goes from G to ; the universal property automatically constructs a map for you. The advantage of using the universal property rather than defining a map out of directly is that you don't repeat the verification that the map is well-defined --- it's been done once and for all in the proof above.
Should you ever need to know how the magic map is defined, refer to the proof (and the commutativity of the diagram).
Remarks.
commutes means that .
Example. ( Using the universal property to construct a group map) To illustrate how the universal property is used, suppose you wanted to construct a homomorphism from the quotient group to . How would you do it?
The universal property tells me to construct a group map from to which contains in its kernel --- that is, which sends to 0. Now consists of all multiples of , so what I'm looking for is a group map which sends to 0.
To ensure that what I get is a group map, I should probably guess a linear function --- something like
If , then . There is no question of {\it solving} this equation for a and b, since there is one equation and two variables. But I just need some a and b that work --- and one "obvious" way to do this is to set and , since
Notice that , would work, too. In fact, there are infinitely many possibilities.
So I define by
It's easy to check that this is a group map, and I constructed it so that . Therefore, the universal property automatically produces a group map . It is defined by
Why not just define the map this way to begin with? If you did, you'd have to check that the map was well-defined. It's less messy to use the universal property to construct the map as above.
Next, I want to prove a powerful result that you can use to construct isomorphisms of quotient groups. The idea is simple, and uses the universal property of the quotient. First, I'll prove a very useful lemma.
Lemma. Let be a group map. is injective if and only if .
Proof. ( ) Suppose is injective. Since , . Conversely, let , so . Then , so by injectivity . Therefore, , so .
( ) Suppose . I want to show that is injective. Suppose . I want to show that .
Hence, , so , and . Therefore, is injective.
Example. ( Proving that a group map is injective) Define by
Prove that f is injective.
As usual, is a group under vector addition. Since I can write f in the form
f is a linear transformation, so it's a group map.
To show f is injective, I'll show that the kernel of f consists of only the identity: . Suppose . Then
Since , I know by linear algebra that the matrix equation has only the trivial solution: . This proves that if , then , so . Since , it follows that .
Hence, f is injective.
Theorem. ( The First Isomorphism Theorem) Let be a group map, and let be the quotient map. There is an isomorphism such that the following diagram commutes:
Proof. Since maps G onto and , the universal property of the quotient yields a map such that the diagram above commutes. Since is onto, so is ; in fact, if , by commutativity .
It remains to show that is injective.
By the previous lemma, it suffices to show that . Since maps out of , the "1" here is the identity element of the group , which is the subgroup . So I need to show that .
However, this follows immediately from commutativity of the diagram. For if and only if . This is equivalent to , or , or --- i.e. .
Example. ( Using the First Isomorphism Theorem to show two groups are isomorphic) Use the First Isomorphism Theorem to prove that
is the group of nonzero real numbers under multiplication. is the group of positive real numbers under multiplication. is the group consisting of 1 and -1 under multiplication (it's isomorphic to ).
I'll define a group map from onto whose kernel is .
Define by
Since
is a group map.
If is a positive real number, then
Therefore, is surjective: .
Finally, clearly sends 1 and -1 to the identity , and those are the only two elements of which map to 1. Therefore, .
By the First Isomorphism Theorem,
Note that I didn't construct a map explicitly; the First Isomorphism Theorem constructs the isomorphism for me.
Example. ( Using the First Isomorphism Theorem to show two groups are isomorphic) Prove that
First, I need to define a group map . I'd like it to be surjective, and I want the kernel to be .
To ensure that the function I define is a group map, I should use a {\it linear} function. So I'm thinking of something of the form .
In order for to be , I want to choose a and b so that . This means I want . An easy way to arrange this is to switch the 3 and the 1, then negate one of the numbers (say the 3). This gives , .
Following this line of thought, I define by
First, I'll show that f is a group map:
To show that , I'll show that each of these sets is contained in the other. First, I'll show . To do this, I'll take an element of and show that it's in . An element of has the form for some , and to be in means that f maps the element to 0. So I do the computation:
This proves that . Hence, .
To show that , I take an element in and show that it's in . Suppose that . Then , so , or . Then
Thus, , so .
Next, I have to show that f is surjective, i.e. that . To do this, I take an arbitrary element and find an input such that . In this situation, you're trying to "solve" for m and n in terms of p, but there are usually many 's which will work.
Now , so this will equal p if and . In other words, . Hence, .
Hence, the First Isomorphism Theorem says that
Here's what this means pictorially. A coset of in consists of the lattice points on the line . The isomorphism essentially maps this coset to the x-intercept .
For example, consider the point . Since
and are in the same coset of .
What is the line through and ? It has slope , so its equation is
The x-intercept is -1 --- and sure enough,
Lemma. If is a surjective group map and , then .
Proof. , so , and .
Let , so . Then
Therefore, is a subgroup.
(Notice that this does not use the fact that K is normal. Hence, I've actually proved that the image of a subgroup is a subgroup.)
Now let , , so . I want to show that . Since is surjective, for some . Then
But because K is normal. Hence, . It follows that is a normal subgroup of H.
Theorem. ( The Second Isomorphism Theorem) Let , . Then
Proof. I'll use the First Isomorphism Theorem. To do this, I need to define a group map .
To define this group map, I'll use the Universal Property of the Quotient.
The quotient map is a group map. By the lemma preceding the Universal Property of the Quotient, . Since , it follows that .
Since is a group map and , the Universal Property of the Quotient implies that there is a group map given by
If , then . Therefore, is surjective.
I claim that .
First, if (so ), then . Since H is the identity in , it follows that .
Conversely, suppose , so
The last equation implies that , so .
Thus, .
By the First Isomorphism Theorem,
There is also a Third Isomorphism Theorem (sometimes called the Modular Isomorphism, or the Noether Isomorphism). It asserts that if and , then
You can prove it using the First Isomorphism Theorem, in a manner similar to that used in the proof of the Second Isomorphism Theorem.
Copyright 2008 by Bruce Ikenaga