proof of first isomorphism theorem
The proof consist of several parts which we will give for completeness. Let denote . The following calculation validates that for every and :
Let be a map that sends the coset to .
Since is defined on representatives we need to show that it is well defined. So, let and be two elements of that belong to the same coset (i.e. ). Then, is an element of and therefore (because is the kernel of ). Now, the rules of homomorphism show that and that is equivalent to which implies the equality .
Next we verify that is a homomorphism. Take two cosets and , then:
Finally, we show that is an isomorphism (i.e. a bijection). The kernel of consists of all cosets in such that but these are exactly the elements that belong to so only the coset is in the kernel of which implies that is an injection. Let be an element of and its pre-image. Then, equals thus and therefore is surjective.
were is the natural projection that takes to . We will conclude by verifying this. Take in then, as needed.
|Title||proof of first isomorphism theorem|
|Date of creation||2013-03-22 12:39:19|
|Last modified on||2013-03-22 12:39:19|
|Last modified by||uriw (288)|