proof of third isomorphism theorem

We’ll give a proof of the third isomorphism theorem using the Fundamental homomorphism theorem.

Let $G$ be a group, and let $K\subseteq H$ be normal subgroups of $G$. Define $p,q$ to be the natural homomorphisms from $G$ to $G/H$, $G/K$ respectively:

$$p(g)=gH,q(g)=gK\forall g\in G.$$

$K$ is a subset of $\ker (p)$, so there exists a unique homomorphism $\phi :G/K\to G/H$ so that $\phi \circ q=p$.

$p$ is surjective, so $\phi$ is surjective as well; hence $\mathrm{im}\phi =G/H$. The kernel of $\phi$ is $\ker (p)/K=H/K$. So by the first isomorphism theorem we have

$$(G/K)/\ker (\phi )=(G/K)/(H/K)\approx \mathrm{im}\phi =G/H.$$

Title	proof of third isomorphism theorem
Canonical name	ProofOfThirdIsomorphismTheorem
Date of creation	2013-03-22 15:35:09
Last modified on	2013-03-22 15:35:09
Owner	Thomas Heye (1234)
Last modified by	Thomas Heye (1234)
Numerical id	5
Author	Thomas Heye (1234)
Entry type	Proof
Classification	msc 20A05