Use the fact that $U(R \times S) = U(R) \times U(S)$, where $U(R)$ denotes the group of units of the ring $R$, to prove that $U (\Bbb{Z}_9) \simeq \Bbb{Z}_6$ and $U(\Bbb{Z}_{15}) \simeq \Bbb{Z}_{4} \times \Bbb{Z}_2$.
I imagine we also have to use the fact that $\Bbb{Z}_{mn} \simeq \Bbb{Z}_m \times \Bbb{Z}_n$, where $gcd(m,n)=1$, is a ring isomorphism. I figured out the second part: $\Bbb{Z}_{15} \simeq \Bbb{Z}_{5} \times \Bbb{Z}_{3}$ which implies $U(\Bbb{Z}_{15}) \simeq U(\Bbb{Z}_{5} \times \Bbb{Z}_{3}) = U (\Bbb{Z}_{5}) \times U(\Bbb{Z}_{3}) \simeq \Bbb{Z}_4 \times \Bbb{Z}_2$. But I am having trouble with the first. I could use a hint. The best I could come up with is $U(\Bbb{Z}_{18}) \simeq U(\Bbb{Z}_2 \times \Bbb{Z}_9) = U(\Bbb{Z}_2) \times U(\Bbb{Z}_9) \simeq U(\Bbb{Z}_9)$, but this would involve arguing that $U(\Bbb{Z}_{18}) \simeq \Bbb{Z}_6$.
Hint:
$\bigl|\,(\mathbf Z/9\mathbf Z)^\times \bigr|=\varphi(9)=6$, and $(\mathbf Z/9\mathbf Z)^\times$ is an abelian group.