Find all the elements k in U(9) such that U(9)=<k>?

Answer 1

The order of U(9) is φ(9) = 2*3 = 6.
Now note that 2 is a primitive root(i.e., generator) of 3.
Since 2² = 4(mod 9), 2 is also a primitive root mod 9.
So the powers of 2 are (mod 9)
2, 4,8,7,5,1.
Next, 4³ = 1(mod 9)
8² = 1(mod 9)
7³ = 1(mod 9)
So the only other generator of U(9) is 5.
Here are the powers of 5 mod 9:
5,7,8, 4, 2,1.
To sum up: 2 and 5 are all the elements k
in U(9) such that U(9) = .