Consider finite groups H⊂GH\subset G. Denote by H∖GH\backslash G the right orbit space. That is:

H∖G={Hσ1,…,Hσn}H\backslash G = \{H\sigma_1,\dots,H\sigma_n\}

for some choice of σ1,…,σn∈G\sigma_1,\dots,\sigma_n \in G. Consider the right action of GG on H∖GH\backslash G. If we fix a g∈Gg\in G, this will decompose H∖GH\backslash G into a number of equivalence classes like so:

{Hσ1,…,Hσ1gf1(g)},…,{Hσn,…,Hσngfn(g)}.\{H\sigma_1,\dots,H\sigma_1g^{f_1(g)}\},\dots,\{H\sigma_n,\dots,H\sigma_ng^{f_n(g)}\}.

Assume that H,GH,G are such that for all g∈Gg\in G, fi(g)f_i(g) is constant in ii. Does this imply that HH is normal in GG?

Note that the converse is true. If HH were normal in GG, then right cosets correspond to left cosets and therefore, we can identify the above right action of GG with the left action of GG in which case fi(g)f_i(g) is simply the order of gg modulo HH.

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Oh, that was very easy. If you just paste your comment as an answer, I will accept it.

– Asvin

2 days ago

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1 Answer

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Yes it’s equivalent. If the subgroup is not normal then choose g∈H∖f−1Hfg \in H \setminus f^{-1}Hf for some f∈Gf \in G. Then the orbit of HH under gg has length 11, but the orbit of HfHf has length greater than 11. There is no need to assume that the groups are finite.i