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Created March 19, 2016 22:55
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mathedit.md

Suppose $\langle a^{d_{1}} \rangle \leq \langle a^{d_{2}} \rangle$. Then

$$a^{d_{1}} = (a^{d_{2}})^{m} \textrm{, for some } m \in \mathbb{Z}$$

Thus

$$ a^{d_{1}-md_{2}} = e = \textrm{ identity.}$$

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