Prove U(14) is cyclic. Best explanation about this topic☝️☝️☝️

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My proof:
Order of U(14) = φ(14) = 14*6/7*1/2 = 6
There are two groups of order 6, the symmetric group S_3 and the cyclic group Z_6.

U(14) is abelian so it must be isomorphic to Z_6.

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very good explation sir by the way what is the name of the book?

RahulRoy-pwcb