Multiplying modulo 20 
For a positive integer $n \ge 2$, define $U(n)= \{k \in \Bbb Z_n $| gcd$(k,n)=1\}$. Then $U(n)$ is a group under multiplication modulo $n$. Find the order of $U(20)$. Is it possible to generate $U(20)$ with a single element? If so, state this element and show it is a generator; if not, find a minimal generating set. 

I found that $U(n)=\{1,3,7,9,11,13,17,19\}$ and |$U(20)$|$=8$. 
I know that $U(20)$ cannot be generated with a single element. So far I have: $<1,3>$ but I know that I need more elements. Multiplying the elements of $U(20)$ $modulo \ 20$ is tedious. Is there a better way, i.e., a shortcut, to do this? 

Edit:
I figured out that my subgroup is $<3,11>$, but I just did it by multiplying and subtracting 20 over and over. Is there a better way to do this than multiplying over and over and then subtracting 20 over and over?
 A: Since you are only required to find a minimal generating set and not a set of minimum size, you can just start with the whole group and throw out elements until you can't throw out any more.  This shouldn't be too hard.  So to start out
$$
U(n) = \langle 1,3,7,9,11,13,17,19 \rangle
$$
Certainly you can throw out $1$.
Then $3 \cdot 3 = 9, 11 \cdot 3 = 13, 11 \cdot 7 = 17$, so throw out $9, 13, 17$ as well.
$$
U(n) = \langle 3,7,11,19 \rangle
$$
Now $7^2 \cdot 11 = 9 \cdot 11 = 19$, and
$3 \cdot 7 = 1$ i.e. $7 = 3^{-1}$.  So
$$
U(n) = \langle 3, 11 \rangle.
$$
A: You are right about the elements of $U(20)$ and the order of the group. Another way to name the elements (here listed in the same order as you do) is:
$$U(20) = \{ 1,3,7,9,-9,-7,-3,-1 \}$$
and that might be easier to multiply (and should not confuse you when you do modulo 20 operations).
Your generating set $\langle 1,3 \rangle$ does not need the element $1$ since $1$ is the neutral element in this group.
To check if the group can be generated by one element, lets first see what $\langle 3 \rangle$ is. $3\cdot 3 = 9$ and $9\cdot 3 = 27 \equiv 7$, and finally $7\cdot 3 = 21 \equiv 1$, so:
$$\langle 3 \rangle = \{ 1,3,9,7 \}$$
a cyclic subgroup of order $4$ whose other generator is $7 \equiv 3^3$:
$$\langle 7 \rangle = \{ 1,7,9,3 \}$$
and of course $\langle 9 \rangle = \{ 1,9 \}$ from this.
With the other elements, we get similarly:
$$\langle -3 \rangle = \{ 1,-3,9,-7 \}$$
and:
$$\langle -7 \rangle = \{ 1,-7,9,-3 \}$$
and $\langle -9 \rangle = \{ 1,-9 \}$ and $\langle -1 \rangle = \{ 1,-1 \}$.
We see that the group is not cyclic.
You will be able to find a set of two generators. Take $3$ as one of them, and try an element not already in $\langle 3 \rangle$ as the other one. Good luck.
A: $|U(n)|=\varphi(n)$ where $\varphi$ is Euler-phi function.
And it is multiplicative,i.e $\varphi(20)=\varphi(4).\varphi(5)=2.4=8$
$U(n)\cong Aut(Z_n)$ thus, $U(20)\cong Aut(Z_{20})\cong (Aut(Z_4)\times Aut(Z_5)\cong Z_2\times Z_4$
Thus,it is not cyclic.
Note that,I used following facts;
$Z_{mn}\cong Z_m\times Z_n$ if and only if $gcd(m,n)=1$ and in that case you can say $Aut(Z_{mn})\cong (Aut(Z_m)\times Aut(Z_n))$
