Systematically list the cyclic subgroups of $U(30)$. 
List the cyclic subgroups of $U(30)$.

$$U(30) = \{1, 7, 11, 13, 17, 19, 23, 29\}$$
Not sure how to systematically approach finding all the cyclic subgroups of $U(30)$.
Attempt:
$1$ is the identity and not a generator.
Taking powers of $7 \pmod {30}$ produces every element in $U(30)$ so $7$ is a generator.
Not sure how to best go from there. I thought because $U(30)$ has order $8$ meaning that there will be $4$ generators (because $\phi(8) = 4$) but my textbook says the solution is $\langle1\rangle, \langle7\rangle,\langle11\rangle,\langle17\rangle,\langle19\rangle,\langle29\rangle$ are the generators? Is this correct? Do I have to manually calculate for each element of $U(30)$?
 A: To list all cyclic subgroups, we note that 


*

*The subgroup generated by $a$, namley, $\langle a\rangle$ is same as the subgroup generated by $a^{-1}$, $\langle a^{-1} \rangle$



*

*The  identity $1$ generates $\langle 1\rangle=\{1\}$

*The inverse of $7$ is $13$, since $7 \cdot 13=91 \equiv 1(\mod 30)$ 
So $\langle 7\rangle=\langle 13\rangle$

*Next we move onto $11$. The inverse of $11$ is $11$ itself, since $11 \cdot 11=121 \equiv 1(\mod 30)$. In this case, we have only one subgroup, $\langle 11\rangle$. Similarly $19$ and $29$ has own inverses. So we get $\langle 19\rangle$ and $\langle 13\rangle$

*Next we move onto $17$.The inverse of $17$ is $23$, since $17 \cdot 23=391 \equiv 1(\mod 30)$. So $\langle 17\rangle=\langle 23\rangle$
In summary, the cyclic subgroups of $U(30)$ are $$\langle 1\rangle,\langle 7\rangle, \langle 11\rangle, \langle 19\rangle, \langle 29\rangle, \langle 17\rangle$$ which is exactly the answer as your book says ! [actually the  book says about subgroups, not generators]
