Given the following Euler groups : $\begin{align*} U_{12} &= \{1,5,7,11\}\\ U_{16} &= \{1,3,5,7,9,11,13,15\} \end{align*}$
I want to prove that they are not cyclic. I used the following theorem :
A group of order $n$ is cyclic if and only if it has an element of order $n$.
Let's take for example $U_{12}$: (I will use the notation of $o(x)$ to denote the order of the element $x\in G$)
$\begin{align*} o(5)&\colon 5^2=25 \to 25\bmod 12 = 1\to o(5)=2\\ o(7)&\colon 7^2= 49 \to 49\bmod 12 = 1 \to o(7)=2\\ o(11) &\colon 11^2 = 121 \to 121\bmod 12=1 \to o(11)=2 \end{align*} $
Then by using the above theorem , this group is indeed not a cyclic group.
Question : do I really have to check each element in the group for its order ?
Regards