Cyclic groups and Unit groups

Question

So I wanted to find the the subgroup $3 \in U(20)$ where $U(20)$ is the unit group of $\mathbb Z_{20}$ (i.e. all integers in $\mathbb Z_{20}$ that are relatively prime to $20$). I thought that this is a cyclic subgroup where $3$ is the generator and $U(20) = \lbrace 1,3,7,9,11,13,17\rbrace$. Now $3$ is a generator for $\mathbb Z_{20}$, so the subgroup is going to be the whole $U(20)$. However, I got confused when the answer said $\lbrace 1,3,7,9\rbrace$. What am I misunderstanding here?

Answer 1

$3$ does not generate $U(20)$.

$$3^2=9\\ 3^3=27\equiv 7\pmod {20}\\ 3^4\equiv 3\cdot 7\pmod {20}\equiv 1\pmod {20}$$

Answer 2

$U(20)$ is a multiplicative group whereas $\mathbb{Z}_{20}$ is an additive group. That is, the group operation on $U(20)$ is multiplication $\mod 20$, while the group operation on $\mathbb{Z}_{20}$ is addition $\mod 20$.

So, the subgroup of $U(20)$ generated by $3$ is: $\{3,3^2=9,3^3=7,3^4=1\}$.

Answer 3

You were wrong to think $3$ is a generator of $U(20)$; just check $$3\cdot3\equiv9\pmod{20};\quad 3\cdot9\equiv7\pmod{20};\quad 3\cdot7\equiv1\pmod{20}.$$ Thus, we clearly see that $\langle 3\rangle=\{1,3,7,9\}$.