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If we know that $a$ is in a group, then any power of $a$ would also be in the group because it is closed under the group operation, right?

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    $\begingroup$ Yep, that's right (including the case of negative exponents) $\endgroup$ – Hagen von Eitzen Mar 22 '15 at 16:22
  • $\begingroup$ You're correct. Moreover, if the group is of finite order, there must exist $i$ and $j$ such that, at some point, $a^i = a^j$. More info here $\endgroup$ – Miguelgondu Mar 22 '15 at 16:30
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    $\begingroup$ Where else would it be? $\endgroup$ – Slade Mar 22 '15 at 22:49
  • $\begingroup$ Yes, for any integer power. $\endgroup$ – paw88789 Mar 22 '15 at 22:59
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A binary operation on a set $S$ is a function $f$ with domain $S\times S$ and co-domain $S$. This means any combination of the group's operation stays in the group.

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If you want a formal proof, $a=a^1$ is in the group by your hypothesis, and assuming that $a^{i-1}$ is in the group it follows that $a^i = a a^{i-1}$ is in the group. So by induction, $a^n$ is in the group for all $n$.

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