# the kernel of a group homomorphism is a normal subgroup

In this entry we show the following simple lemma:

###### Lemma 1.

Let $G$ and $H$ be groups (with group operations^{} ${\mathrm{\ast}}_{G}$, ${\mathrm{\ast}}_{H}$ and identity elements^{} ${e}_{G}$ and ${e}_{H}$, respectively) and let $\mathrm{\Phi}\mathrm{:}G\mathrm{\to}H$ be a group homomorphism^{}. Then, the kernel of $\mathrm{\Phi}$, i.e.

$$\mathrm{Ker}(\mathrm{\Phi})=\{g\in G:\mathrm{\Phi}(g)={e}_{H}\},$$ |

is a normal subgroup^{} of $G$.

###### Proof.

Let $G,H$ and $\mathrm{\Phi}$ be as in the statement of the lemma and let $g\in G$ and $k\in \mathrm{Ker}(\mathrm{\Phi})$. Then, $\mathrm{\Phi}(k)={e}_{H}$ by definition and:

$\mathrm{\Phi}(g{\ast}_{G}k{\ast}_{G}{g}^{-1})$ | $=$ | $\mathrm{\Phi}(g){\ast}_{H}\mathrm{\Phi}(k){\ast}_{H}\mathrm{\Phi}({g}^{-1})$ | ||

$=$ | $\mathrm{\Phi}(g){\ast}_{H}({e}_{H}){\ast}_{H}\mathrm{\Phi}({g}^{-1})$ | |||

$=$ | $\mathrm{\Phi}(g){\ast}_{H}\mathrm{\Phi}({g}^{-1})$ | |||

$=$ | $\mathrm{\Phi}(g){\ast}_{H}\mathrm{\Phi}{(g)}^{-1}$ | |||

$=$ | ${e}_{H},$ |

where we have used several times the properties of group homomorphisms and the properties of the identity element ${e}_{H}$. Thus, $\mathrm{\Phi}(gk{g}^{-1})={e}_{H}$ and $gk{g}^{-1}\in G$ is also an element of the kernel of $\mathrm{\Phi}$. Since $g\in G$ and $k\in \mathrm{Ker}(\mathrm{\Phi})$ were arbitrary, it follows that $\mathrm{Ker}(\mathrm{\Phi})$ is normal in $G$. ∎

###### Lemma 2.

Let $G$ be a group and let $K$ be a normal subgroup of $G$. Then there exists a group homomorphism $\mathrm{\Phi}\mathrm{:}G\mathrm{\to}H$, for some group $H$, such that the kernel of $\mathrm{\Phi}$ is precisely $K$.

###### Proof.

Simply set $H$ equal to the quotient group^{} $G/K$ and define $\mathrm{\Phi}:G\to G/K$ to be the natural projection^{} from $G$ to $G/K$ (i.e. $\mathrm{\Phi}$ sends $g\in G$ to the coset $gK$). Then it is clear that the kernel of $\mathrm{\Phi}$ is precisely formed by those elements of $K$.
∎

Although the first lemma is very simple, it is very useful when one tries to prove that a subgroup^{} is normal.

###### Example.

Let $F$ be a field. Let us prove that the special linear group^{} $\mathrm{SL}(n,F)$ is normal inside the general linear group^{} $\mathrm{GL}(n,F)$, for all $n\ge 1$. By the lemmas, it suffices to construct a homomorphism^{} of $\mathrm{GL}(n,F)$ with $\mathrm{SL}(n,F)$ as kernel. The determinant^{} of matrices is the homomorphism we are looking for. Indeed:

$$det:\mathrm{GL}(n,F)\to {F}^{\times}$$ |

is a group homomorphism from $\mathrm{GL}(n,F)$ to the multiplicative group^{} ${F}^{\times}$ and, by definition, the kernel is precisely $\mathrm{SL}(n,F)$, i.e. the matrices with determinant $=1$. Hence, $\mathrm{SL}(n,F)$ is normal in $\mathrm{GL}(n,F)$.

Title | the kernel of a group homomorphism is a normal subgroup |
---|---|

Canonical name | TheKernelOfAGroupHomomorphismIsANormalSubgroup |

Date of creation | 2013-03-22 17:20:34 |

Last modified on | 2013-03-22 17:20:34 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 8 |

Author | alozano (2414) |

Entry type | Theorem |

Classification | msc 20A05 |

Related topic | KernelOfAGroupHomomorphism |

Related topic | NaturalProjection |