cyclic group of order m

**apple2009** · November 19th 2009, 04:57 AM

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.

**Drexel28** · November 19th 2009, 06:36 AM

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 $r$ is unique. If $r|m$ what can we say about $a^{\frac{m}{r}}$ ?

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cyclic group of order m

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