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A cyclic group of order 60 has

1. 12 Generators
2. 15 Generators
3. 16 Generators
4. 20 Generators

## 2 Answers

Best answer

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 !

One way is to note that φ(d) is also equal to the number of possible generators of the cyclic group

### 4 Comments

If I check each positive number less than 60 which are co prime of 60, I get following numbers :

1,2,11,13,17,19,21,23,29,31,37,41,43,47,53,59. These are 15 in numbers.. Which one I am missing
What about 7 ??
I found my mistake.. I missed 7 and 49 and added 2 which should not be there

@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.

References:

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