Third 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
Statement
Suppose is a group, and and are Normal subgroup (?)s of , such that . Then we have the following natural isomorphism:
Where the isomorphism sends a coset in to the coset in .
Note that this statement makes sense at the level of a group isomorphism only when both and are normal in . Otherwise, the statement is still true at the level of sets, but we cannot make sense of it as a group isomorphism.
Related facts
Other isomorphism theorems
- First isomorphism theorem
- Second isomorphism theorem
- Fourth isomorphism theorem (also known as the lattice isomorphism theorem or the correspondence theorem)
- Zassenhaus isomorphism theorem
Facts used
- Normality satisfies intermediate subgroup condition
- Normality is image-closed (not really required; implicitly shown in the proof)
- First isomorphism theorem
Proof
Given: A group , with normal subgroups and , such that
To prove:
Proof: Note first that all the three expressions for quotient groups make sense. and make sense because are normal in . Moreover, since normality satisfies intermediate subgroup condition, is also normal in .
Next, observe that is a normal subgroup in , because normality is image-closed: under the quotient map by , the normal subgroup of gets sent to a normal subgroup of . Thus, the left side makes sense.
Let's now describe the isomorphism from the left side to the right side:
In other words, the map takes a coset of and gives the corresponding coset of . This is well-defined, because if , then , so .
Further, the map is a homomorphism. For this, observe that it sends the identity element to the identity element, preserves the group multiplication, and preserves the inverse map.
Further, the map is surjective, because any coset occurs as the image of under .
Finally, we need to determine the kernel of the map. This is given by the set of such that . This is precisely those cosets of that are in , which is the same as the coset space . Hence, the kernel of the map is precisely .
Thus, the surjective homomorphism has kernel precisely . By the first isomorphism theorem, we get:
.
References
Textbook references
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Theorem 19, Section 3,3, Page 98
- Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, ^{More info}, Exercise 8, Miscellaneous Problems, Page 236