Trying to prove that a group is not Cyclic 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
 A: 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 $n$. 
A: Some shortcuts are available. For example, when testing $3$ in $U_{16}$, you find its powers are $3,9,11,1$, so you don't have to test $9$ or $11$. Do you see why?
A: Well really the "theorem" is the definition of cyclic group...it is a group generated by one element.
Now if I give you a group and you check that all but one element isn't a generator, how do you know that the one you didn't check isn't a generator?
So yes, you must check ALL elements.
A: What of if your set is infinite? I think the best idea is to find the general form of the elements in the set (if possible) and pick arbitarary element, check your assertion and then generalised. But in your case, since the elements are finite, you have to check it for all the elements in the set.
