# normality of subgroups of prime index

###### Proposition.

If $H$ is a subgroup^{} of a finite group^{} $G$ of index $p$, where $p$ is the smallest prime dividing the order of $G$, then $H$ is normal in $G$.

###### Proof.

Suppose $H\le G$ with $|G|$ finite and $|G:H|=p$, where $p$ is the smallest prime divisor^{} of $|G|$, let $G$ act on the set $L$ of left cosets^{} of $H$ in $G$ by left , and let $\phi :G\to {S}_{p}$ be the http://planetmath.org/node/3820homomorphism^{} induced by this action. Now, if $g\in \mathrm{ker}\phi $, then $gxH=xH$ for each $x\in G$, and in particular, $gH=H$, whence $g\in H$. Thus $K=\mathrm{ker}\phi $ is a normal subgroup^{} of $H$ (being contained in $H$ and normal in $G$). By the First Isomorphism Theorem^{}, $G/K$ is isomorphic to a subgroup of ${S}_{p}$, and consequently $|G/K|=|G:K|$ must http://planetmath.org/node/923divide $p!$; moreover, any divisor of $|G:K|$ must also $|G|=|G:K||K|$, and because $p$ is the smallest divisor of $|G|$ different from $1$, the only possibilities are $|G:K|=p$ or $|G:K|=1$. But $|G:K|=|G:H||H:K|=p|H:K|\ge p$, which $|G:K|=p$, and consequently $|H:K|=1$, so that $H=K$, from which it follows that $H$ is normal in $G$.
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Title | normality of subgroups of prime index |
---|---|

Canonical name | NormalityOfSubgroupsOfPrimeIndex |

Date of creation | 2013-03-22 17:26:38 |

Last modified on | 2013-03-22 17:26:38 |

Owner | azdbacks4234 (14155) |

Last modified by | azdbacks4234 (14155) |

Numerical id | 13 |

Author | azdbacks4234 (14155) |

Entry type | Theorem |

Classification | msc 20A05 |

Related topic | Coset |

Related topic | GroupAction |

Related topic | ASubgroupOfIndex2IsNormal |