# 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 $\mathrm{ker}(p)$, 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 $\mathrm{ker}(p)/K=H/K$. So by the first isomorphism theorem^{} we have

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

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 |