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I can't understand why the statement in the title is true. This is what I know so far that is relevant:

- A subgroup of a cyclic group G = <g> is cyclic and is <g^k> for some nonnegative integer k. If G is finite (say |G|=n) then k can be chosen so that k divides n, and so order of g^k is n/k.

- Let G = <g> be a cyclic group. If g has infinite order, then G is isomorphic to the integers with addition. If o(g) = n, then G is isomorphic to Z/nZ with addition.

And then apparently the last point implies the statement in my title and I have no idea how!

Thanks,

Matt

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# For every positive integer n there is a unique cyclic group of order n

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