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# proof of first isomorphism theorem

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

$\begin{array}[]{llll}f(gkg^{{-1}})&=&f(g)\,f(k)\,f(g)^{{-1}}&\text{($f$ is an % homomorphism)}\\ &=&f(g)\,1_{H}\,f(g)^{{-1}}&\text{(definition of $K$)}\\ &=&1_{H}&\end{array}$ |

Hence, $gkg^{{-1}}$ is in $K$. Therefore, $K$ is a normal subgroup of $G$ and $G/K$ is well-defined.

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

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

Since $\theta$ is defined on representatives we need to show that it 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, the rules of homomorphism show that $f(g_{1})^{{-1}}f(g_{2})=1$ and that is equivalent 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:

$\begin{array}[]{llll}\theta(g_{1}K\cdot g_{2}K)&=&\theta(g_{1}g_{2}K)&\text{(% operation in $G/K$)}\\ &=&f(g_{1}g_{2})&\text{(definition of $\theta$)}\\ &=&f(g_{1})f(g_{2})&\text{($f$ is an homomorphism)}\\ &=&\theta(g_{1}K)\theta(g_{2}K)&\text{(definition of $\theta$)}\end{array}$ |

Finally, we show that $\theta$ is an isomorphism (i.e. a bijection). 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 $\operatorname{Im}f$ 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 that the following diagram is commutative:

$\xymatrix{G\ar[rd]_{{f}}\ar[r]^{{\pi}}&G/K\ar[d]_{{\theta}}\\ &H}$ |

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.

## Mathematics Subject Classification

20A05*no label found*

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