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    Fields, Elements and Cyclic Groups

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    Find H K in {see attachment}, if H = <|3|> and K = <|5|>.

    This is all the problem says. I know the answer, but I do not know the reasoning.

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    Z16* is the set of Z16, without the element [0] and the operation figured out is the multiplication.
    That means the set contains the elements [1], [2],...[15].
    <|3|> is the cyclic group generated by the element [3]
    This group is:
    H = {[1], [3], [9], ...

    Solution Summary

    Fields, Elements and Cyclic Groups are investigated. The solution is detailed and well presented.