Sachchidanand Prasad

29-03-2023

Problem: Find out the number of order $10$ elements in $\left( \mathbb{Z}_{41}^{\star} ,\times \right) $, where $\mathbb{Z} _{41}^\star $ is the set of all invertible elements in $\mathbb{Z} _{41}$.

Solution: We recall a theorem.

If $G$ is a cyclic group of order $n$ and if $k$ is a positive integer that divides $n$, then the number of elements of order $k$ equals to $\varphi (k)$ which is the number of positive integers less than $k$ and coprime to $k$.

Since $41$ is a prime number, so all non-zero elements in $\mathbb{Z} _{41}$ are invertible. Therefore, the order of the group is $40$. Now $10$ divides $40$ so using the above theorem, the number of order $10 $ elements will be \[ \varphi (10) = \varphi (5\times 2) = \varphi (5) \times \varphi (2) = 4 \times 1 = 4. \]

Here we used the properties of the $\varphi $ function (this is also known as Euler-phi function ).
  1. $\varphi $ is multiplicative, that is if $p$ and $q$ are coprime, then $\varphi (pq) = \varphi (p) \varphi (q)$.
  2. For a prime $p$, the number of coprime positive integers less than $p$ is $p-1$ and hence, $\varphi (p) = p-1$.