First isomorphism theorem
This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article is about an isomorphism theorem in group theory.
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Name
This result is termed the first isomorphism theorem, or sometimes the fundamental theorem of homomorphisms.
Statement
General version
Let be a group and be a homomorphism of groups. The first isomorphism theorem states that the kernel of is a Normal subgroup (?), say , and there is a natural isomorphism:
where denotes the image in of under .
More explicitly, if is the quotient map, then there is a unique isomorphism such that .
Version for surjective homomorphism
This is a special case of the more general statement.
Let be a group and be a surjective homomorphism of groups. Then, if is the kernel of , we have:
More explicitly, if is the quotient map, then there is a unique isomorphism such that .
Universal algebraic statement
- In the variety of groups, every ideal (normal subgroup) is a kernel
- In the variety of groups, a congruence is completely determined by its kernel. In other words, simply knowing the inverse image of the identity element for a surjective homomorphism, determines the nature of the homomorphism.
This is encoded by saying that the variety of groups is ideal-determined.
Related facts
Related facts about groups
- Normal subgroup equals kernel of homomorphism
- Second isomorphism theorem
- Third isomorphism theorem
- Fourth isomorphism theorem
Facts used
- Normal subgroup equals kernel of homomorphism: Given any homomorphism of groups, the kernel of (i.e., the inverse image of the identity element) is a normal subgroup of . Further, given any normal subgroup of , there is a natural quotient group .
Proof
Given: A homomorphism of groups , with kernel (i.e. is the inverse image of the identity element).
To prove: is a normal subgroup, and
Proof: Two steps of the proof are done at normal subgroup equals kernel of homomorphism (fact (1)):
- The kernel of any homomorphism is a normal subgroup
- If is a normal subgroup, we can define a quotient group which is the set of cosets of , with multiplication of cosets given by:
It now remains to show that we can identify isomorphically with . Consider the map from to :
We first argue that this map is well-defined. For this, observe that if , then:
In other words, any two elements in the same coset of in get mapped to the same element of .
Next, we argue that the map is a group homomorphism. Indeed, if and are two cosets, then:
(similar checks work for identity element and inverses).
Next, we argue that the map is injective. Indeed, if is sent to the identity element, then , forcing .
Finally, we argue that the map is surjective. By definition, any element in can be written as for some , and hence occurs as .
References
Textbook references
- Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, ^{More info}, Page 68-69, Theorem (10.9)
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 97 (Theorem 16): the proof is spread across previous sections, and is not given after the statement