A cyclic group of order 60 has
Ans -> C
No of generators in Group (Cyclic group) too is given by Euler's_totient_function, i.e. no of elements less than N & Co prime to N.
Here $60 = 2^2 * 3 * 5$
No of generators possible are $= 60 ( 1 - 1/2) (1 - 1/3) (1-1/5) = 60 * 1/2 * 2/3 * 4/5 = 16.$
So total $16$ Generators !
Reference -> http://math.stackexchange.com/questions/786452/how-to-find-a-generator-of-a-cyclic-group
One way is to note that φ(d) is also equal to the number of possible generators of the cyclic group
-> https://en.wikipedia.org/wiki/Euler's_totient_function
@anchitjindal07
$7$ and $49$ both should be added because these are relatively prime to $60$.
In $\mathbb{Z_{60}},$ the generators are the numbers $0,\cdots,59$ that are relatively prime to $60.$ These are $1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59.$ We can write Multipliatively , $G=\langle{a\rangle}$ where the order of $a$ is $60$, the generators are $a^{1},a^{7},a^{11},a^{13},a^{17},a^{19},a^{23},a^{29},a^{31},a^{37},a^{41},a^{43},a^{47},a^{49},a^{53},a^{59}.$
Two positive integers are said to be relatively prime if their greatest common divisor is $1.$ For instance, $10$ and $7$ are relatively prime as they share no factors other than $1.$
$\implies$Note that neither integers need to be prime in order for them to be relatively prime $8$ and 15 are both not prime, yet they are relatively prime.
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