# proof that $\mathrm{exp}G$ divides $|G|$

The following is a proof that $\mathrm{exp}G$ divides $|G|$ for every finite group^{} $G$.

###### Proof.

By the division algorithm^{}, there exist $q,r\in \mathbb{Z}$ with $$ such that $|G|=q(\mathrm{exp}G)+r$. Let $g\in G$. Then ${e}_{G}={g}^{|G|}={g}^{q(\mathrm{exp}G)+r}={g}^{q(\mathrm{exp}G)}{g}^{r}={({g}^{\mathrm{exp}G})}^{q}{g}^{r}={({e}_{G})}^{q}{g}^{r}={e}_{G}{g}^{r}={g}^{r}$. Thus, for every $g\in G$, ${g}^{r}={e}_{G}$. By the definition of exponent, $r$ cannot be positive. Thus, $r=0$. It follows that $\mathrm{exp}G$ divides $|G|$.
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Title | proof that $\mathrm{exp}G$ divides $|G|$ |
---|---|

Canonical name | ProofThatoperatornameexpGDividesG |

Date of creation | 2013-03-22 13:30:32 |

Last modified on | 2013-03-22 13:30:32 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 10 |

Author | Wkbj79 (1863) |

Entry type | Proof |

Classification | msc 20D99 |