Show that the group is cyclic.

Question

I'm trying to show that the group $U(Z_{54})$ is cyclic.

To start, I found the divisors of 54 = {1, 2, 3, 6, 9, 18, 27, 54}

Then I started to find the elements using the powers of a.

Where $U(Z_{54})= <a>$

Therefore, $<a^{54/1}>$={ e }

$ <a^{54/2}>$= { e, $ a^{36}$ }

and such, where the last would be < $a^{54/54}$> = { e }

The things I would like to know is, how does this process show that it's cyclic? Is it because the first divisor (1) and the last divisor (54) is equal to e? Therefore creating a cycle?

Also, what would the < a > to the powers finding the elements process be called?

Thank you!

Answer 1

Hint: (1). Suppose $s$ and $t$ are relatively prime. Then $U(st) \cong U(s) \oplus U(t).$ [ Notation: $U(n) := U(\mathbb{Z}_n)$ ].

(2). $|U(n)| = \phi (n),$ where $\phi$ is the Euler function.

(3). If $G$ and $H$ are two finite cyclic group of orders $m, n$ respectively, with gcd$(m, n) = 1$, then $G \oplus H$ is cyclic.

Note: For this particular case, may be it is easier to compute every element of $U(54)$ and to show that it is actually a cyclic group. But the above mentioned results will enable you to handle a wide range of problems of this kind.

Answer 2

A way to know that is by thinking the problem through number theory. The order of $U( \mathbb{Z}_{54})$ is $\phi(54) = \phi(2) \cdot \phi(3^{3}) = 3^{3} - 3^{2} = 18 = 2 \cdot 3^{2}$. As its order is of the form $2 p^{\alpha}$ then primitive roots exists and we are done.

Hope this helps!!