Normal subgroups

#matematik-groups

Definition: Normal subgroup

Let be a group and be a subgroup. Then is a normal subgroup if for every , , i.e. for every , , and we write

Remark

Let be a group and be a subgroup. Then the following are equivalent:

  2. For every ,
  3. For every ,
  4. For every ,

Proof:
(i)(iv): Clear since .
(i)(ii): Clear since .
(iii)(iv): We have meaning and . Since the same applies for , we are done.
(ii)(iii): Since every is bijective, we have (ii)
for every ,
for every ,
for every , since is a group.

Theorem: Kernels are normal subgroups

Let be a homomorphism. Then .

Proof:
For every and every we have so .

Remark
  • For any group , the trivial subgroup is normal.
  • If is abelian, then every subgroup is normal since .
Theorem

Let be a group and . Then

  1. for every .
  2. is a group with respect to the multiplication in a), with unit element .
  3. The canonical projection defined by for every , is a surjective group homomorphism with .

Proof:
a) Since is normal, we have and , so b) For any we have , so inverses exist.
c) It is surjective by definition. It is a homomorphism since . Finally, we have since if then .

From this we see that there is a strong correlation between kernels and normal subgroups; every kernel is a normal subgroup and every normal subgroup is the kernel of some homomorphism.

Theorem: Universal property of quotient groups

Let be a group homomorphism and such that . Then there is a unique group homomorphism such that . Moreover,

  4. is injective if and only if .

Proof:
Define by . This is well-defined since if then , so . This also shows uniqueness of , since it must satisfy .
a) Clear by definition of .
b) .
c) Follows from b).
d) We know is injective iff , iff .

Corollary

If is a surjective group homomorphism, then is canonically isomorphic to .

Theorem: First isomorphism theorem

Let be a group, , and . Then:

  4. The canonical homomorphism is an isomorphism.

Proof:
c) Clear because .
b) Since is normal we have which implies so is closed under composition, and we also have , so , so is closed under inverses, meaning .
a) We construct a homomorphism by which is surjective and .
d) From the universal property we get a bijection .

Theorem: Second isomorphism theorem

Let be a group, and such that . Then

  3. The canonical homomorphism is an isomorphism.

Proof:
a) Clear since .
b) We define as . This is well defined since if then so . is also surjective, and so .
c) From the universal property we get a bijection .