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