If G=<a> is a cyclic group of order m generated by an element a∈G, then for each divisor r of m exhibit explicitly an element c∈G of order r.

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- Nov 19th 2009, 04:57 AMapple2009cyclic group of order m
If G=<a> is a cyclic group of order m generated by an element a∈G, then for each divisor r of m exhibit explicitly an element c∈G of order r.

- Nov 19th 2009, 06:36 AMDrexel28
Have you even tried this? This really isn't too hard. It doesn't even ask you to show that the element you find with order $\displaystyle r$ is unique. If $\displaystyle r|m$ what can we say about $\displaystyle a^{\frac{m}{r}}$?