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How do I find the order m of each element $\sigma $ and find the subgroup { $e,\sigma,\sigma^{2},···,\sigma^{m−1}$ } generated by such element?

LaurenceMeister
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  • Recall that element order divides group order. There aren't that many divisors of $|S_3|$ ;) – Radost Waszkiewicz Oct 28 '20 at 13:37
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    The phrasing of your Question could be clearer. Are you interested in one specific element $\sigma \in S_3$ or in all of them? Please include the context of your problem, such as where the problem occurs or what approach you've already explored. – hardmath Oct 28 '20 at 13:43
  • The order of the group of permuatation of $n$ letters is $n!$ by Langrange theorem we know that if $H$ is a subgroup then the order of $H$ divides $n!$ so you need to know the diviseur of $n!$ – Aster Phoenix Oct 28 '20 at 13:48
  • @hardmath I am interested in all of them, I did mis phrase the question, it is edited to be more feasible now. – LaurenceMeister Oct 28 '20 at 14:35
  • Are you able to explicitly write down an element of $S_3$ (as a permutation), then find its successive powers (as composed permutations)? – Eric Towers Oct 28 '20 at 14:41
  • @EricTowers Well I know that S3 can be separated into: {e,(1,2), (2,3), (1,3), (1,2,3), (1,3,2)} however I don't know how I would go about finding its successive powers. – LaurenceMeister Oct 28 '20 at 14:54
  • You seem to be saying "powers of the group." No. Pick an element. Find its successive powers. Do you know how to find the product of permutations? – Eric Towers Oct 28 '20 at 14:55
  • @erictowers Well using the elements listed above, if I singled out say (1,3), I know it's product in the context of the subgroup would be {e, (1,3), $(1,3)^2$,..., $(1,3)^m-1$} but I don't know how to expand such products. – LaurenceMeister Oct 28 '20 at 15:07
  • See https://math.stackexchange.com/questions/31763/multiplication-in-permutation-groups-written-in-cyclic-notation – Eric Towers Oct 28 '20 at 15:08
  • @EricTowers that's perfect, thank you! – LaurenceMeister Oct 28 '20 at 15:33
  • I'm glad to help! – Eric Towers Oct 28 '20 at 15:34

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