Finding the generator of cyclic group $U(Z_{27})$

Question

$U(\mathbb{Z}_{27})$ is a group of order $18$.

$U(\mathbb{Z}_{27})=\{1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26\}$

How do I find the generators to prove that this group is cyclic?

The final aim is to prove that group $U(\mathbb{Z}_{54})$ is cyclic using $U(\mathbb{Z}_{54})\cong U(\mathbb{Z}_{27})$.

Answer 1

Hint: In general, if $g$ is a generator for $U(p)$ then $g$ or $g+p$ is a generator for $U(p^n)$ for all $n$. Finding a generator for $U(3)$ is easy...

Answer 2

In general, for an odd $p$, if $g$ is a generator of $U(p)$, one of $g$ or $g+p$ is not of order $p-1$ in $U(p^2)$, nor can it be of order $p$ because of Little Fermat, so that it is a generator of U(p^2). You can prove inductively that it is a generator of $U(p^k)$ for any $k\ge 2$ by proving it can't be of order $p^i$ or $p^i(p-1)$ for $i<k-1$.

Answer 3

$U(\mathbb Z_{27}) = \{1,2,4,5,7,8,10,11,13,14,16,17,19,20,23,22,25,26 \}.$ Claim is that it is a cyclic group (with respect to multiplication) and generated by $2.$

$2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32 = 5, 2 \cdot 5 = 10, 4 \cdot 5 = 20, 8 \cdot 5 = 40 = 13, 2 \cdot 13 = 26, 4 \cdot 13 = 52 = 25, 2 \cdot 25 = 50 = 23, 2 \cdot 23 = 46 = 19, \cdots$ (I think you can finish it from here.)

Note: You don't need to check for all the powers of $2.$ The idea is that if $x, y \in U(\mathbb Z_{27})$ are powers of $2,$ then $xy$ is also a power of $2.$