# proof of first isomorphism theorem

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

$$\begin{array}{cccc}f(gk{g}^{-1})\hfill & =\hfill & f(g)f(k)f{(g)}^{-1}\hfill & \text{(}f\text{is an homomorphism)}\hfill \\ & =\hfill & f(g){\mathrm{\hspace{0.17em}1}}_{H}f{(g)}^{-1}\hfill & \text{(definition of}K\text{)}\hfill \\ & =\hfill & {1}_{H}\hfill & \end{array}$$ |

Hence, $gk{g}^{-1}$ 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 $\theta :G/K\to \mathrm{Im}f$ be a map that sends the coset $gK$ to $f(g)$.

Since $\theta $ is defined on representatives we need to show 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:

$$\begin{array}{cccc}\theta ({g}_{1}K\cdot {g}_{2}K)\hfill & =\hfill & \theta ({g}_{1}{g}_{2}K)\hfill & \text{(operation in}G/K\text{)}\hfill \\ & =\hfill & f({g}_{1}{g}_{2})\hfill & \text{(definition of}\theta \text{)}\hfill \\ & =\hfill & f({g}_{1})f({g}_{2})\hfill & \text{(}f\text{is an homomorphism)}\hfill \\ & =\hfill & \theta ({g}_{1}K)\theta ({g}_{2}K)\hfill & \text{(definition of}\theta \text{)}\hfill \end{array}$$ |

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 $\mathrm{Im}f$ 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 that
the following diagram is commutative^{}:

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

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.

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 |