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

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

###### Proof.

By the division algorithm, there exist $q,r\in{\mathbb{Z}}$ with $0\leq r<\operatorname{exp}~{}G$ such that $|G|=q(\operatorname{exp}~{}G)+r$. Let $g\in G$. Then $e_{G}=g^{|G|}=g^{q(\operatorname{exp}~{}G)+r}=g^{q(\operatorname{exp}~{}G)}g^{% r}=(g^{\operatorname{exp}~{}G})^{q}g^{r}=(e_{G})^{q}g^{r}=e_{G}g^{r}=g^{r}$. 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 $\operatorname{exp}~{}G$ divides $|G|$. ∎

Title proof that $\operatorname{exp}~{}G$ divides $|G|$ ProofThatoperatornameexpGDividesG 2013-03-22 13:30:32 2013-03-22 13:30:32 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Proof msc 20D99