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

The following is a proof that, for every group $G$ that has an exponent and for every $g\in G$, $|g|$ divides $\mathrm{exp}G$.

###### Proof.

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

Canonical name | ProofThatgDividesoperatornameexpG |

Date of creation | 2013-03-22 13:30:35 |

Last modified on | 2013-03-22 13:30:35 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 8 |

Author | Wkbj79 (1863) |

Entry type | Proof |

Classification | msc 20D99 |