We can write U225⋍U15×U15 [By chinese remainder theorem]
Hence any element of U15 has orders 1,2 or 4. Since ∣U15∣=φ(15)=8
there is no element of U15 that generates it, hence U15 is not a cyclic group.
So, we can write:
U(225)=U(152)=U(15)=1,2,4,7,8,11,13,14.The cyclic subgroups generated by these elements are:<2>=<8>=1,2,4,8,<4>=1,4,<7>=1,4,7,13,<11>=1,11(note that 11=−4),<14>=1,14(note that 14=−1),<1>=1.Observe that U(15) cannot be cyclic since it has threesubgroups of order 2 and two of order 4.
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