Second 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.
View a complete list of isomorphism theorems| Read a survey article about the isomorphism theorems
Name
This result is termed the second isomorphism theorem or the diamond isomorphism theorem (the latter name arises because of the diamond-like shape that can be used to describe the theorem).
Statement
General statement
Suppose is a group, and are two subgroups of such that normalizes ; in other words for every . Then, is a group, with (i.e., is a normal subgroup of ), and (i.e. is a normal subgroup of ), and:
Particular case
Suppose is a group, is a normal subgroup, and an arbitrary subgroup, such that . Then:
Definitions used
Term | Relevant definitions(s) |
---|---|
normal subgroup | A subgroup of a group is termed a normal subgroup if for any and , . Equivalently, is normal in if and only if for any , . |
isomorphism of groups | Given groups and a map is termed an isomorphism of groups if is bijective and for all . |
quotient group | For a normal subgroup of a group , the quotient group is the group whose elements are the cosets , and where the product of two cosets is the unique coset comprising the products of elements in the two cosets. |
Particular cases
Specific possibilities for the relationship between and
Note that for each of these, if we have , then the statement made for applies to .
Situation | Interpretation |
---|---|
is trivial | In this case, and are actually the same thing, the isomorphism is the identity map, and the result gives no information. |
is trivial | In this case, and are both trivial groups with the trivial isomorphism between them. Again, the result gives no information. |
is trivial | In this case, is an internal semidirect product with normal piece and complement (and hence retract) . The result says that , i.e., complement to normal subgroup is isomorphic to quotient. |
is trivial, both and normalize each other | In this case, is an internal direct product of and . We get and . |
Specific possibilities for and the location of and within
Nature of | Nature/parameters for and | Interpretation of and | Interpretation of and | Fact in this context that corresponds to second isomorphism theorem |
---|---|---|---|---|
group of integers | , | If and , then and | which is cyclic of order . which is cyclic of order | The arithmetic fact that , i.e., the product of two numbers equals the product of their gcd and lcm. |
vector space of dimension | is a subspace of dimension , is a subspace of dimension | has dimension and has dimension ; these dimensions are not determined by | The dimension of is and the dimension of is | The isomorphism between these (which turns out to also be a vector space isomorphism) preserves dimension, and we get or . This is a vector space analogue of the inclusion-exclusion principle. |
is an external direct product of groups | There are subsets of such that is the subgroup where all coordinates outside are the identity, and is the subgroup where all coordinates outside are the identity. | corresponds to all coordinates outside being the identity, and corresponds to . | corresponds to (in a loose subgroup-quotient identification) and to . | The fact that . In finite cardinality terms, . |
Related facts
Facts about normal subgroups
- Normality satisfies transfer condition: If are subgroups and is normal in , then is normal in .
- Normality satisfies intermediate subgroup condition: If are groups and is normal in , then is normal in .
General version of the result
- Product formula: The set-theoretic version of the product formula establishes a bijection which is the same as the bijection of the second isomorphism theorem, but without the conditions of normality. The bijection is purely at the set-theoretic level.
Facts used
Proof
Given: A group , subgroups such that for all .
To prove: is a group, is normal in , is normal in , and .
Proof:
is a subgroup of
Condition | In symbols | Justification |
---|---|---|
Closure under multiplication | For , want to show | Observe that for some , by the condition, so . |
Identity element | The identity element of is in | The identity element is in both and , so it is in . |
Inverses | Given , | . Thus, the inverse of any element in is also in . |
is normal in
Observe that, by the condition, for any . Thus, for and , we have . Thus, is normal in .
is normal in
We now prove that is normal in . Pick . Then such is a subgroup. Also, since for any , we have , so . Thus, , and we get that is normal in .
Definition of isomorphism and proof that it works
Finally, define the isomorphism between and as follows:
.
We check that this satisfies all the required conditions:
Condition | Verification |
---|---|
Sends cosets to cosets | Note first that if two elements are in the same coset of , they are in the same coset of . Thus, the map sends cosets of to cosets of . (This is fact (1)). |
Well-defined with specified domain and co-domain | Further, if , then . In other words, if the original coset is in , the new coset is in . Thus, the map is well-defined from to . |
Injective | Suppose . That means that , forcing . But we anyway have , so , forcing that and are in the same coset of . Thus, . |
Surjective | Any left coset of in can be written as where . Thus, we can write where . Then, , with . Thus, . |
Homomorphism | Suppose . Then, . Thus, the map is a homomorphism. |
References
Textbook references
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, Page 97, Theorem 18 of Section 3.3, ^{More info}
- Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, Page 236, Exercise 7 of Miscellaneous Problems, ^{More info}