# 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 $\operatorname{exp}~{}G$ ProofThatgDividesoperatornameexpG 2013-03-22 13:30:35 2013-03-22 13:30:35 Wkbj79 (1863) Wkbj79 (1863) 8 Wkbj79 (1863) Proof msc 20D99