proof of first isomorphism theorem


The proof consist of several parts which we will give for completeness. Let K denote kerf. The following calculation validates that for every gG and kK:

f(gkg-1)=f(g)f(k)f(g)-1(f is an homomorphism)=f(g) 1Hf(g)-1(definition of K)=1H

Hence, gkg-1 is in K. Therefore, K is a normal subgroupMathworldPlanetmath of G and G/K is well-defined.

To prove the theoremMathworldPlanetmath we will define a map from G/K to the image of f and show that it is a function, a homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath and finally an isomorphismMathworldPlanetmathPlanetmath.

Let θ:G/KImf be a map that sends the coset gK to f(g).

Since θ is defined on representatives we need to show that it is well defined. So, let g1 and g2 be two elements of G that belong to the same coset (i.e. g1K=g2K). Then, g1-1g2 is an element of K and therefore f(g1-1g2)=1 (because K is the kernel of G). Now, the rules of homomorphism show that f(g1)-1f(g2)=1 and that is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to f(g1)=f(g2) which implies the equality θ(g1K)=θ(g2K).

Next we verify that θ is a homomorphism. Take two cosets g1K and g2K, then:

θ(g1Kg2K)=θ(g1g2K)(operation in G/K)=f(g1g2)(definition of θ)=f(g1)f(g2)(f is an homomorphism)=θ(g1K)θ(g2K)(definition of θ)

Finally, we show that θ is an isomorphism (i.e. a bijection). The kernel of θ consists of all cosets gK in G/K such that f(g)=1 but these are exactly the elements g that belong to K so only the coset K is in the kernel of θ which implies that θ is an injection. Let h be an element of Imf and g its pre-image. Then, θ(gK) equals f(g) thus θ(gK)=h and therefore θ is surjectivePlanetmathPlanetmath.

The theorem is proved. Some version of the theorem also states that the following diagram is commutativePlanetmathPlanetmathPlanetmath:

\xymatrixG\ar[rd]f\ar[r]π&G/K\ar[d]θ&H

were π is the natural projectionMathworldPlanetmath that takes gG to gK. We will conclude by verifying this. Take g in G then, θ(π(g))=θ(gK)=f(g) 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