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

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Solution:

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], ...

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