Sachchidanand Prasad

03-05-2023

Problem: Let $G$ be a non-cyclic group of order $95$. Find out the number of elements in $G$ of order $5$.

Solution: The group of order $95$ can not be cyclic. We have the following result.

Let $p$ and $q$ be two primes with $p \lt q$. If $p$ does not divide $q-1$, then any group of order $pq$ is cyclic.

Since, $95 = 19 \times 5$ and $5$ does not divide $19-1=18$, therefore, there does not exist any cyclic group of order $95$.