computationally efficient way to determine the cube of a U-group is the U-group itself.

Question

Question:

Show that $U\left ( 55 \right )^{3}=\left \{ x^{3} \mid x \in U\left ( 55 \right ) \right \}$ is $U\left ( 55 \right )$.

A laborious way to do this is to determine the order of $U\left ( 55 \right )$. In fact, the order is 36. Taking the cube of each element under mod 55 would suffice. But is there a more efficient way around this?

Thanks in advance.

Answer 1

$|U(55)| = \phi(55) = \phi(5\times11) = 4\times10 = 40$

Since $\gcd(3,40)=1$, the mapping $a \mapsto a^3$ is an automorphism.