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

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# 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 $\operatorname{exp}~{}G$. ∎

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## Mathematics Subject Classification

20D99*no label found*

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