In group theory, two groups are said to be isomorphic if there exists a bijective homomorphism (also called an isomorphism) between them. An isomorphism between two groups and means (informally) that and are the same group, written in two different ways.
Many groups that come from quotient constructions are isomorphic to groups that are constructed in a more direct and simple way. There are three standard isomorphism theorems that are often useful to prove facts about quotient groups and their subgroups.
This theorem is the most commonly used of the three. Given a homomorphism between two groups, the first isomorphism theorem gives a construction of an induced isomorphism between two related groups.
Let and be two groups and let be a group homomorphism. Then the kernel is a normal subgroup of and
Here is the image of the homomorphism which is a subgroup of The isomorphism is completely natural: letting sends an element of the quotient to the image of under That is,
Let be the group of nonzero complex numbers under multiplication. Consider the homomorphism given by Then the first isomorphism theorem says that
Now and (that is, is onto), so the conclusion is that i.e. is isomorphic to a nontrivial quotient of itself.
What is wrong with this argument?
The second isomorphism theorem relates two quotient groups involving products and intersections of subgroups.
Let be a group, let be a subgroup, and let be a normal subgroup. Then is a subgroup of and
Let be the symmetric group on three letters. Let be the subgroup generated by the transposition and let be the subgroup generated by the transposition
Then the second isomorphism theorem gives an isomorphism where
Now so the right side has order So the left side has order Now and so But Lagrange's theorem says that must divide which is
What has gone wrong with this argument?
I. As defined above, is not a subgroup of so Lagrange's theorem does not apply.
II. The order of a quotient of finite groups is not always equal to
III. is not normal, so the second isomorphism theorem does not apply.
The third isomorphism theorem is extremely useful in analyzing the normal subgroups of a quotient group.
Let be a group, and a normal subgroup. Then
(1) There is a natural one-to-one correspondence between subgroups of and subgroups of containing The correspondence matches a subgroup of containing with the subgroup of
(2) This correspondence preserves normality: the normal subgroups of correspond to the normal subgroups of containing That is, a subgroup of containing is normal in if and only if is normal in
(3) The correspondence preserves quotients: if is a normal subgroup of containing then
Consider the group of invertible matrices. The determinant is a homomorphism from to the multiplicative group of nonzero real numbers. It is surjective as well (for any the determinant of is ). Then the first isomorphism theorem says that where is the kernel of the determinant homomorphism. Here, is simply the set of matrices which the homomorphism sends to the identity of i.e. the subgroup of matrices with determinant This is called So by the first isomorphism theorem.
A nontrivial consequence of the third isomorphism theorem is that any subgroup of containing is normal, because the quotient group is abelian, so all its subgroups are normal. The third isomorphism theorem also gives a full description of the subgroups of containing ; they consist of the matrices whose determinants lie in a subgroup of
Suppose are positive integers with Then Find the order of the quotient
This is an application of the third isomorphism theorem. Let Then Now and so
Prove that using the second isomorphism theorem and the previous example.
Let The group law is addition, so the second isomorphism theorem becomes Now , since it is the set of multiples of both and
On the other hand, is the set of all integers of the form A well-known result in elementary number theory known as Bezout's identity says that this set consists of the multiples of So Taking the orders of both sides and using the previous example gives and the result follows.
Proof of the first isomorphism theorem:
To prove the first theorem, we first need to make sure that is a normal subgroup (where is the kernel of the homomorphism , the set of all elements that get mapped to the identity element of the target group ). This means we need to prove that for any homomorphism, the identity element is in the kernel, the inverse of elements in the kernel are also in the kernel, the product of two elements in the kernel is also in the kernel, and for any
To see that the identity element of is an element of , consider , where denotes the identity element of . On one hand:
but on the other hand,
so we must have
We similarly prove the existence of inverse in : if , then
Closure is easy to prove:
To prove that is normal within , note that for any and any ,
The proof that is also a group can be accomplished by similar means. For simplicity's sake, let's refer to as and as . The quotient group
is defined as the group of cosets , with multiplication given by . To prove the isomorphism between this group and , consider the homomorphism
By the definition of the image of a function, any function is surjective, or onto, into its image, so we just need to prove that is injective, or one-to-one:
Start with the equation
Left-multiply both sides by to get:
This means that
for some , and
, and .
Because of the closure of under multiplication, every element of the form will also be in . But these are precisely all the elements of . So . And the second relationship shows us that . So we must have . This completes the proof of the first isomorphism theorem.
Proof of the second isomorphism theorem:
This proof relies on the first isomorphism theorem. It is easy to see that is normal within and is normal within . Since both sides of the isomorphism are quotient groups, it would be nice if we could construct homomorphisms with , and with , then prove that . Fortunately, one of the many equivalent definitions of a normal subgroup is that there exists a homomorphism whose kernel is that subgroup. Let be the homomorphism from to some group which satisfies this property for . All the elements of can be expressed as . Notice that:
, and this will be different from the identity element if and only if . This means
Now consider the same homomorphism applied to the subgroup of . Its kernel will be all the elements of which are also elements of , or . Its image when it's applied to this group is defined as , which is equal to the image of when it's applied to . So the two quotient groups
are both isomorphic to the same group, . Therefore they are isomorphic to one another.