Disproving that $U_{20}$ is cyclic 
Let $U_{20}$ be the group of invertible elements in $(\mathbb Z_{20}, \cdot)$. Prove that $U_{20}$ is not cyclic, as in it is not generated by one element.

To prove that a group is cyclic I just find some element that generates it. But what is the way of proving that a group is not cyclic? What do we look for?
I thought about this:

$U_{20} = \{1,3,7,9,11,13,17,19\}$, then from a theory if it must be cyclic then it must be isomorphic to some $\mathbb Z_n$. 

Am I in the right direction?
 A: Since $U_{20} = \{1, 3, 7, 9, 11, 13, 17, 19\}$ has only eight elements, it is feasible to check this by hand: Computing the powers of $3$ gives that $\langle 3 \rangle = \{1, 3, 7, 9\}$, so if the group is cyclic, it is not generated by any of $1, 3, 7, 9$ (since the groups they generate are subgroups of $\langle 3 \rangle \lneq U_{20}$. Similarly, $\langle 11 \rangle = \{1, 9, 13, 17\}$, and so neither $13$ nor $17$ generate $U_{20}$. The only elements unaccounted for are $11$ and $19$, but both of these have order $2$, so neither of these generate the group either. (In fact, we can conclude from the fact that $U_{20}$ has elements of order $4$ but no element of order $8$ that $(U_{20}, \cdot) \cong (\Bbb Z_4, +) \times (\Bbb Z_2, +)$.)
A: In a cyclic group $G$ of even order, the set $\{ x \in G : x^2=1 \}$ has exactly $2$ elements, but in $U_{20}$ it has $4$ elements: $1,9,11,19$.
More generally, $U(4m)$ is never cyclic because $(2m\pm1)^2=4m^2\pm 4m+1\equiv 1 \bmod 4m$. And of course $1$ and $m-1$ satisfy $x^2=1$.
A: If $g$ generates $U_{20}$ we must have a sequence $g^2,g^4,g^8=e$ of three distinct squares. But we have
$$
\begin{array}{|c|c|}
\hline
x&1&3&7&9&11&13&17&19\\
\hline
x^2&1&9&9&1&1&9&9&1\\
\hline
\end{array}
$$
and it is quite clear, that no such sequence exists.
