# a subgroup of index 2 is normal

###### Lemma.

Let $\mathrm{(}G\mathrm{,}\mathrm{\cdot}\mathrm{)}$ be a group and let $H$ be a subgroup^{} of $G$ of index 2. Then $H$ is normal in $G$.

###### Proof.

Let $G$ be a group and let $H$ be an index 2 subgroup of $G$. By definition of index, there are only two left cosets^{} of $H$ in $G$, namely:

$$H,{g}_{1}H$$ |

where ${g}_{1}$ is any element of $G$ which is not in $H$. Notice that if ${g}_{1},{g}_{2}$ are two elements in $G$ which are not in $H$ then ${g}_{1}\cdot {g}_{2}$ belongs to $H$. Indeed, the coset ${g}_{1}{g}_{2}H\ne {g}_{1}H$ (because ${g}_{1}{g}_{2}={g}_{1}h$ would immediately yield ${g}_{2}=h\in H$) and so ${g}_{1}{g}_{2}H=H$ and ${g}_{1}{g}_{2}\in H$.

Let $h\in H$ be an arbitrary element of $H$ and let $g\in G$. If $g\in H$ then $gh{g}^{-1}\in H$ and we are done. Otherwise, assume that $g\notin H$. Thus $gh\notin H$ and by the remark above $gh{g}^{-1}=(gh){g}^{-1}\in H$, as desired. ∎

Title | a subgroup of index 2 is normal |
---|---|

Canonical name | ASubgroupOfIndex2IsNormal |

Date of creation | 2013-03-22 15:09:25 |

Last modified on | 2013-03-22 15:09:25 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 5 |

Author | alozano (2414) |

Entry type | Theorem |

Classification | msc 20A05 |

Related topic | Coset |

Related topic | QuotientGroup |

Related topic | NormalityOfSubgroupsOfPrimeIndex |