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proof that  divides 
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(Proof)
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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 $0 \le r<|g|$ such that $\operatorname{exp}~G=q|g|+r$ Since $e_G=g^{\operatorname{exp}~G}=g^{q|g|+r}=(g^{|g|})^qg^r=(e_G)^qg^r=e_Gg^r=g^r$ 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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"proof that  divides  " is owned by Wkbj79.
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Cross-references: positive, order, division algorithm, divides, exponent, group, proof
This is version 5 of proof that  divides  , born on 2003-03-11, modified 2007-05-30.
Object id is 4092, canonical name is ProofThatGDividesExpG.
Accessed 1626 times total.
Classification:
| AMS MSC: | 20D99 (Group theory and generalizations :: Abstract finite groups :: Miscellaneous) |
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Pending Errata and Addenda
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