G is a group acting on S. H is a subgroup of G and let s be anything in S. Is it true that the order of s under action of G is divisible by the order of s under action of H?
Orbit of a subgroup
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group-theory
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0The stabilizer $H_s$ of $s$ in $H$ is exactly the intersection $G_s \cap H$. So your statement is reduced to ask if $[H \cap G:H \cap G_s] / [G:G_s]$. Any ideas on how to go further? – 2011-05-16
1 Answers
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If G is the symmetric group on n letters, then the orbit of s has exactly n elements. Saying that H ≤ G means that H is a permutation group on n letters. Does every permutation group on n letters have orbits that are divisors of n?
Perhaps check n = 3 to start. A similar investigation n = 5 should also be instructive.