CYCLIC GROUP Problem | TIFR 201O | PART A | PROBLEM 1

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