Try this problem from TIFR GS-2010 which involves the concept of Cyclic Group.
A cyclic group of order $60$ has
CYCLIC GROUP
ORDER OF AN ELEMENT
EULER'S PHI FUNCTION
Answer:$16$ generators
TIFR 2010|PART A |PROBLEM 1
CONTEMPORARY ABSTRACT ALGEBRA: JOSEPH GALLIAN
If G be a cyclic group of order n ,then the number of generators of G is $\phi(n)$
Let us define Euler's phi function:-
$\phi(n)$=the number of positive integers less than $n$ and prime to $n$
i)If $n$ is prime then $\phi(n)=n-1$
ii)If $m$,$n$ be two integers which are relatively prime $\phi{mn}=\phi(m)\phi(n)$
iii)If $n$ be a prime and $k$ be any positive integer,$\phi(p^k)=p^k(1-\frac{1}{p})$
Now $60$ can be written as product of $2^2$,$3$,$5$
Therefore $\phi(60)=\phi(2^2).\phi(3).\phi(5)$
Now $\phi(2^2)=2$.......i
Now $\phi(3)=(3-1)=2$ ........ii
Now $\phi(5)=(5-1)=4$ ........iii
Now multiply i,ii &iii and get the result
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