proof of first isomorphism theorem

The proof consist of several parts which we will give forcompleteness. Let $K$ denote $\ker f$. The following calculationvalidates that for every $g\in G$ and $k\in K$:

Hence, $gk{g}^{-1}$ is in $K$. Therefore, $K$ is a normal subgroupof $G$ and $G/K$ is well-defined.

To prove the theorem we will define a map from $G/K$ to the imageof $f$ and show that it is a function, a homomorphism and finallyan isomorphism.

Let $\theta :G/K\to \mathrm{Im}f$ be a map that sends the coset$gK$ to $f(g)$.

Since $\theta$ is defined on representatives we need to show thatit is well defined. So, let $g_1$ and $g_2$ be two elements of $G$that belong to the same coset (i.e. $g_{1}K=g_{2}K$). Then,$g_1^{-1}{g}_2$ is an element of $K$ and therefore$f(g_1^{-1}{g}_2)=1$ (because $K$ is the kernel of $G$). Now, therules of homomorphism show that $f{(g_1)}^{-1}f(g_2)=1$ and that isequivalent to $f(g_1)=f(g_2)$ which implies the equality$\theta (g_{1}K)=\theta (g_{2}K)$.

Next we verify that $\theta$ is a homomorphism. Take two cosets$g_{1}K$ and $g_{2}K$, then:

Finally, we show that $\theta$ is an isomorphism (i.e. abijection). The kernel of $\theta$ 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$\theta$ which implies that $\theta$ is an injection. Let $h$be an element of $\mathrm{Im}f$ and $g$ its pre-image. Then,$\theta (gK)$ equals $f(g)$ thus $\theta (gK)=h$ and therefore$\theta$ is surjective.

The theorem is proved. Some version of the theorem also states thatthe following diagram is commutative:

were $\pi$ is the natural projection that takes $g\in G$ to $gK$.We will conclude by verifying this. Take $g$ in $G$ then,$\theta (\pi (g))=\theta (gK)=f(g)$ as needed.