proof of fourth isomorphism theorem
First we must prove that the map defined by is a bijection. Let denote this map, so that . Suppose , then for any we have for some , and so . Hence , and similarly , so and is injective. Now suppose is a subgroup of and by . Then is a subgroup of containing and , proving that is bijective.
Now we move to the given properties:
If then trivially , and the converse follows from the fact that is bijective.
To show we need only show that if or then . The other cases are dealt with using the fact that . So suppose then clearly because . Similarly for . Similarly, to show we need only show that if or then . So suppose , then for some , giving and so . Similarly for .
Suppose , then for some and since , . Therefore and , and so meaning . Now suppose . Then for some , giving and so . Similarly , therefore and .
Suppose . Then for any we have and so .
Conversely suppose . Consider , the composition of the map from onto and the map from onto . iff which occurs iff therefore for some . However is contained in , so this statement is equivalnet to saying . So is the kernel of a homomorphism, hence is a normal subgroup of .
|Title||proof of fourth isomorphism theorem|
|Date of creation||2013-03-22 14:17:38|
|Last modified on||2013-03-22 14:17:38|
|Last modified by||aoh45 (5079)|