proof that $|g|$ divides $\operatorname{exp}~{}G$

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

Proof.

By the division algorithm, there exist $q,r\in{\mathbb{Z}}$ with $0\leq r<|g|$ such that $\operatorname{exp}~{}G=q|g|+r$ . Since $e_{G}=g^{\operatorname{exp}~{}G}=g^{q|g|+r}=(g^{|g|})^{q}g^{r}=(e_{G})^{q}g^{r% }=e_{G}g^{r}=g^{r}$ , by definition of the order of an element, $r$ cannot be positive. Thus, $r=0$ . It follows that $|g|$ divides $\operatorname{exp}~{}G$ . ∎

Title	proof that $\|g\|$ divides $\operatorname{exp}~{}G$
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