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 :
Hence, is in . Therefore, is a normal subgroup
of and is well-defined.
To prove the theorem we will define a map from to the image
of and show that it is a function, a homomorphism
and finally
an isomorphism
.
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.
The theorem is proved. Some version of the theorem also states that
the following diagram is commutative:
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 |
---|---|
Canonical name | ProofOfFirstIsomorphismTheorem |
Date of creation | 2013-03-22 12:39:19 |
Last modified on | 2013-03-22 12:39:19 |
Owner | uriw (288) |
Last modified by | uriw (288) |
Numerical id | 9 |
Author | uriw (288) |
Entry type | Proof |
Classification | msc 20A05 |