Sachchidanand Prasad

05-04-2023

Problem: Let $U(20)$ be the set of all invertible elements in $\left( \mathbb{Z} _{20}, \times \right) $. It is known that it is a group with multiplication. Find the number of elements of all possible orders.

Solution: By the Lagrange's theorem we know that the order of an element divides the order of the group. Note that the group is $U(20)$ and the number of elements in $U(20)$ is $\varphi (20) = 8$. So the possible order of an element will be divisors of $8$, that is, $1,2,4$ and $8$.

  • The only element of order $1$ in $U(20)$ is the identity element. So the number of elements of order $1$ in $U(20)$ is $1$.
  • The elements of order $2$ in $U(20)$ are $\bar{9}, \bar{11}$ and $\bar{19}$ . So the number of elements of order $2$ in $U(20)$ is $3$.
  • The elements of order $4$ in $U(20)$ are $\bar{3}, \bar{7}, \bar{13}$ and $\bar{17}$ . So the number of elements of order $4$ in $U(20)$ is $4$.
  • The number of elements of order $8$ is zero.
The group $U(n)$ is cyclic if and only if $n=1,2,4,p^l,2p^l$, where $p$ is an odd prime number and $l>0$.