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I know that S(5) is not cyclic.

If $S(5) = (1 2 3 4 5)$, you could say $(1 2 3 4 5) = (2 3)(3 4)(4 5)(1 5) = (1 5)(1 4)(1 3)(1 2)$, giving us various sets of four 2-cycles.

Is 4 the smallest number of elements needed to generate S(5)? Is there a different way to generate $S(5)$ using a smaller number of elements?

Michael Rozenberg
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  • @InterstellarProbe, no, I need to show the smallest number of elements needed to generate S(5) in a proof-like manner, not just for one specific case –  Feb 25 '20 at 19:10
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    Since $S_5$ is not cyclic, it is not generated by one element. Showing that it can be generated by 2 elements shows that you found the generator with the smallest number of elements. So, yes, you do need to show for one specific case. – SlipEternal Feb 25 '20 at 19:14
  • But, if you want, this answer should work for you, as well: https://math.stackexchange.com/questions/520615/12-and-123-dots-n-are-generators-of-s-n – SlipEternal Feb 25 '20 at 19:16

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