The * First Isomorphism Theorem* helps identify
quotient groups as "known" or "familiar" groups.

I'll begin by proving a useful lemma.

* Proposition.* 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. I can write f in the form

Since f has been represented as multiplication by a constant matrix, it 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 surjective, 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

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.* is a group under componentwise addition and
is a group under addition. Let

Prove that .

Define by

Note that

Since f can be expressed as multiplication by a constant matrix, it's a linear transformation, and hence a group map.

Let . Then

Therefore, , and hence .

Let . Then

Hence,

Therefore, . Hence, .

Let . Note that

Hence, .

Thus,

* Example.* is a group under
componentwise addition and is a group under addition. Prove that

Define by

f can be represented by matrix multiplication:

Hence, it's a group map.

Let . Then

Thus, .

Let . Then

Now but . By Euclid's lemma, . Say . Then

Therefore,

Thus, .

Hence, .

Let . Note that

Multiplying by z, I get

Then

This proves that .

Hence,

* Example.* is a group under
componentwise addition. Consider the subgroup

Prove that .

( is a group under componentwise addition.)

Define by

Note that

Since f is defined by matrix multiplication, it is a linear transformation. Hence, it's a group map.

Let . Then

Hence, , and .

Let . Then

Equating the first components, I have , so . Equating the second components, I have , so . Thus,

Therefore, , and so .

Let . Then

Hence, .

Thus,

The first equality follows from . The isomorphism follows from the First Isomorphism Theorem. The second equality follows from .

* Proposition.* 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 2018 by Bruce Ikenaga