<s>[INST] Show that $U_{14}\cong U_{18}$? Because $|U_{14}|=|U_{18}|$ and both of them is cyclic and commutative, so i just need define a function $f : U_{14}\to U_{14}$ that f is homomorphism and bijective.
$f(1)=1, f(3)=5, f(5)=7, f(9)=11, f(11)=13, f(13)=17$
Is that function homomorphism and bijective? If not, what is the function?
 [/INST] Since
\begin{align}
\varphi(14)
&=14\left(1-\frac{1}{2}\right)\left(1-\frac{1}{7}\right)=6 \\[6px]
\varphi(18)
&=18\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=6
\end{align}
we know the two groups can be isomorphic.
If we consider $3\in U_{14}$ and its powers, we get
$$
1=3^0,\quad
3=3^1,\quad
9=3^2,\quad
13=-1=3^3
$$
so the order of $3$ is six.
Similarly, if we consider $5$ in $U_{18}$, we get
$$
1=5^0,\quad
5=5^1,\quad
7=5^2,\quad
17=-1=5^3
$$
so the order of $5$ is six.
Since the two groups are cyclic of the same order, they're isomorphic. You find all isomorphisms by sending one generator of $U_{14}$ (for instance $3$) to any generator of $U_{18}$; the number is $\varphi(6)=2$, so the generators are $5$ and $5^{-1}=11$. The isomorphisms are
$$
f(1)=1, f(3)=5, f(9)=7, f(13)=17, f(11)=13, f(5)=11
$$
and
$$
g(1)=1, g(3)=11, g(9)=13, g(13)=17, g(11)=7, g(5)=5
$$
where $3^k$ is sent to $5^k$ or to $11^k$ respectively.
</s>