Compute the order of a cyclic subgroup in Z/(54)

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.10
What is the order of $\overline{30}$ in $\mathbb{Z}/(54)$? Write out all of the elements and their orders in $\langle \overline{30} \rangle$.

Solution: We know that $$|\overline{30}| = 54/\mathsf{gcd}(30,54) = 54/6 = 9.$$ The elements of $\langle \overline{30} \rangle$ are $$\{ \overline{30}, \overline{6}, \overline{36}, \overline{12}, \overline{42}, \overline{18}, \overline{48}, \overline{24}, \overline{0} \}.$$ The order of each element in $\langle \overline{30} \rangle$ is the same as its order in $\mathbb{Z}/(54)$. Thus we have the following. $$|\overline{30}| = 9$$ $$|\overline{6}| = 54/\mathsf{gcd}(6,54) = 54/6 = 9$$ $$|\overline{36}| = 54/\mathsf{gcd}(36,54) = 54/18 = 3$$ $$|\overline{12}| = 54/\mathsf{gcd}(12,54) = 54/6 = 9$$ $$|\overline{42}| = 54/\mathsf{gcd}(42,54) = 54/6 = 9$$ $$|\overline{18}| = 54/\mathsf{gcd}(18,54) = 54/18 = 3$$ $$|\overline{48}| = 54/\mathsf{gcd}(48,54) = 54/6 = 9$$ $$|\overline{24}| = 54/\mathsf{gcd}(24,54) = 54/6 = 9$$ $$|\overline{0}| = 1$$