Second isomorphism theorem
This article gives the statement, and possibly proof, of a basic fact in group theory.
View a complete list of basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
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
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).
Suppose is a group, is a normal subgroup, and an arbitrary subgroup, such that . Then:
|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.|
Specific possibilities for the relationship between and
Note that for each of these, if we have , then the statement made for applies to .
|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, .|
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.
Given: A group , subgroups such that for all .
To prove: is a group, is normal in , is normal in , and .
is a subgroup of
|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:
|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.|