# fundamental homomorphism theorem

The following theorem is also true for rings (with ideals instead of normal subgroups^{}) or modules (with submodules^{} instead of normal subgroups).

###### theorem 1.

Let $G\mathrm{,}H$ be groups, $f\mathrm{:}G\mathrm{\to}H$ a homomorphism^{}, and let $N$ be a normal subgroup of $G$ contained in $\mathrm{ker}\mathit{}\mathrm{(}f\mathrm{)}$. Then there exists a unique homomorphism $h\mathrm{:}G\mathrm{/}N\mathrm{\to}H$ so that $h\mathrm{\circ}\phi \mathrm{=}f$, where $\phi $ denotes the canonical homomorphism from $G$ to $G\mathrm{/}N$.

Furthermore, if $f$ is onto, then so is $h$; and if $\mathrm{ker}\mathit{}\mathrm{(}f\mathrm{)}\mathrm{=}N$, then $h$ is injective^{}.

###### Proof.

We’ll first show the uniqueness. Let ${h}_{1},{h}_{2}:G/N\to H$ functions such that ${h}_{1}\circ \phi ={h}_{2}\circ \phi $. For an element $y$ in $G/N$ there exists an element $x$ in $G$ such that $\phi (x)=y$, so we have

$${h}_{1}(y)=({h}_{1}\circ \phi )(x)=({h}_{2}\circ \phi )(x)={h}_{2}(y)$$ |

for all $y\in G/N$, thus ${h}_{1}={h}_{2}$.

Now we define $h:G/N\to H,h(gN)=f(g)\forall g\in G$. We must check that the definition is of the given representative; so let $gN=kN$, or $k\in gN$. Since $N$ is a subset of $\mathrm{ker}(f)$, ${g}^{-1}k\in N$ implies ${g}^{-1}k\in \mathrm{ker}(f)$, hence $f(g)=f(k)$. Clearly $h\circ \phi =f$.

Since $x\in \mathrm{ker}(f)$ if and only if $h(xN)={1}_{H}$, we have

$$\mathrm{ker}(h)=\{xN\mid x\in \mathrm{ker}(f)\}=\mathrm{ker}(f)/N.$$ |

∎

A consequence of this is: If $f:G\to H$ is onto with $\mathrm{ker}(f)=N$, then $G/N$ and $H$ are isomorphic.

Title | fundamental homomorphism theorem^{} |
---|---|

Canonical name | FundamentalHomomorphismTheorem |

Date of creation | 2013-03-22 15:35:06 |

Last modified on | 2013-03-22 15:35:06 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 9 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 20A05 |