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  • 4 weeks later...
Posted

I assume you're talking about the cyclic group [math]C_{60}=\{0,1,...,59\}[/math]. Then the generators are exactly the [math]a\in C_{60}[/math] with [math]gcd(a,60)=1[/math]. Thus there are

[math]\varphi(60)=\varphi(3)\varphi(4)\varphi(5)=2\cdot2\cdot4=16[/math] generators.

Posted

I assume you're talking about the cyclic group [math]C_{60}=\{0,1,...,59\}[/math]. Then the generators are exactly the [math]a\in C_{60}[/math] with [math]gcd(a,60)=1[/math]. Thus there are

[math]\varphi(60)=\varphi(3)\varphi(4)\varphi(5)=2\cdot2\cdot4=16[/math] generators.

 

That would be a reasonable assumption, since if a group is generated by any single element it is, by definition cyclic and there is, up to isomorphism, only one such group of a given order.

 

The alternative interpretation of the question would be to determine the elements in each minimal generating set for each group of order 60. This is a much more involved project as one would have to classify all finite groups of order 60 -- I suppose that Sylow theorems would be a good place to start. The project does not appeal to me.

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