TITLE: Trying to prove that a group is not Cyclic
QUESTION [5 upvotes]: 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

REPLY [6 votes]: As I mentioned yesterday in reply to a question of yours: a cyclic group has at most one element of order $2$. 
You accepted that answer. Didn't you read it before accepting it?
In your first two lines, you've shown that $5$ and $7$ both have order $2$ in $U_{12}$. Therefore, the group cannot be cyclic.
Similarly, since $9^2\equiv 1\pmod{16}$ and $15^2\equiv 1\pmod{16}$, $U_{16}$ has at least two distinct elements of order $2$, and therefore cannot be cyclic.
In general, a cyclic group of order $n$ has exactly $\phi(d)$ elements of order $d$ for each divisor $d$ of $n$.