# correspondence between normal subgroups and homomorphic images

Assume, that $G$ and $H$ are groups. If $f:G\to H$ is a group homomorphism^{}, then the first isomorphism theorem^{} states, that the function $F:G/\mathrm{ker}(f)\to \mathrm{im}(f)$ defined by $F(g\mathrm{ker}(f))=f(g)$ is a well-defined group isomorphism. Note that $\mathrm{ker}(f)$ is always normal in $G$.

This leads to the following question: is there a correspondence between normal subgroups^{} of $G$ and homomorphic images^{} of $G$? We will try to answer this question, but before that, let us introduce some notion.

First of all, homomorphic image $\mathrm{im}(f)$ is not only a subgroup^{} of $H$. Actually homomorphic image contains also some data about homomorphism^{}. This observation leads to the following definition:

Definition. Let $G$ be a group. Pair $(H,f)$ is called a homomorphic image of $G$ iff $H$ is a group and $f:G\to H$ is a surjective^{} group homomorphism. We will say that two homomorphic images $(H,f)$ and $({H}^{\prime},{f}^{\prime})$ of $G$ are isomorphic (or equivalent^{}), if there exists a group isomorphism $F:H\to {H}^{\prime}$ such that $F\circ f={f}^{\prime}$.

It is easy to see, that this isomorphism relation^{} is actually an equivalence relation and thus we may speak about isomorphism classes of homomorphic images (which will be denoted by $[H,f]$ for homomorphic image $(H,f)$). Furthermore, if $N\subset G$ is a normal subgroup, then $(G/N,{\pi}_{N})$ is a homomorphic image, where ${\pi}_{N}:G\to G/N$ is a projection^{}, i.e. ${\pi}_{N}(g)=gN$. Let

$$\mathrm{norm}(G)=\{N\subseteq G|N\text{is normal subgroup}\};$$ |

$$\mathrm{h}.\mathrm{im}(G)=\{[H,f]|(H,f)\text{is a homomorphic image of}G\}.$$ |

Proposition^{}. Function $T:\mathrm{norm}(G)\to \mathrm{h}.\mathrm{im}(G)$ defined by $T(N)=[G/N,{\pi}_{N}]$ is a bijection.

Proof. First, we will show, that $T$ is onto. Let $(H,f)$ be a homomorphic image of $G$. Let $N=\mathrm{ker}(f)$. Then (due to the first isomorphism theorem), there exists a group isomorphism $F:G/N\to H$ defined by $F(gN)=f(g)$. This shows, that

$$f(g)=F(gN)=F({\pi}_{N}(g))=(F\circ {\pi}_{N})(g)$$ |

and thus $(G/N,{\pi}_{N})$ is isomorphic to $(H,f)$. Therefore

$$T(N)=[G/N,{\pi}_{N}]=[H,f],$$ |

which completes^{} this part.

Now assume, that $T(N)=T({N}^{\prime})$ for some normal subgroups $N,{N}^{\prime}\in \mathrm{norm}(G)$. This means, that $(G/N,{\pi}_{N})$ and $(G/{N}^{\prime},{\pi}_{{N}^{\prime}})$ are isomorphic, i.e. there exists a group isomorphism $F:G/N\to G/{N}^{\prime}$ such that $F\circ {\pi}_{N}={\pi}_{{N}^{\prime}}$. Let $x\in {N}^{\prime}=\mathrm{ker}({\pi}_{{N}^{\prime}})$ and denote by $e\in G/{N}^{\prime}$ the neutral element. Then, we have

$$e={\pi}_{{N}^{\prime}}(x)=F({\pi}_{N}(x))$$ |

and (since $F$ is an isomorphism) this is if and only if $x\in \mathrm{ker}({\pi}_{N})=N$. Thus, we’ve shown that ${N}^{\prime}\subseteq N$. Analogously (after considering ${F}^{-1}$) we have that $N\subseteq {N}^{\prime}$. Therefore $N={N}^{\prime}$, which shows, that $T$ is injective^{}. This completes the proof. $\mathrm{\square}$

Title | correspondence between normal subgroups and homomorphic images |
---|---|

Canonical name | CorrespondenceBetweenNormalSubgroupsAndHomomorphicImages |

Date of creation | 2013-03-22 19:07:11 |

Last modified on | 2013-03-22 19:07:11 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 20A05 |

Classification | msc 13A15 |

Related topic | HomomorphicImageOfGroup |