Difference between revisions of "Fee subgroup"

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(Created page with "A fee subgroup \(H\) of a linearly ordered subgroup \(G\)is a subgroup such that \(g^{-1}Hg \subset H\) for all \(g > 1\). When \(a(g) = a(h)\) implies \(a(fg) = a(fh)\) and...")
 
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When \(a(g) = a(h)\) implies \(a(fg) = a(fh)\) and \(a(gf) = a(hf)\) for all \(f > 1\),
 
When \(a(g) = a(h)\) implies \(a(fg) = a(fh)\) and \(a(gf) = a(hf)\) for all \(f > 1\),
\(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup<ref name="a">https://googology.fandom.com/ja/wiki/%E3%83%A6%E3%83%BC%E3%82%B6%E3%83%BC%E3%83%96%E3%83%AD%E3%82%B0:Natsugoh/%E5%AE%9F%E5%A4%89%E6%95%B0%E9%96%A2%E6%95%B0%E3%82%92%E5%86%99%E5%83%8F%E3%81%AE%E5%8F%8D%E5%BE%A9%E3%81%8B%E5%90%A6%E3%81%8B%E5%88%A4%E5%AE%9A</ref>.
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\(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup<ref name="a">https://googology.fandom.com/wiki/User_blog:Natsugoh/Discriminating_whether_a_function_of_a_real_variable_is_an_iterated_function</ref>.
  
 
A fee subgroup is not always normal <ref name="b">https://googology.fandom.com/ja/wiki/%E3%83%A6%E3%83%BC%E3%82%B6%E3%83%BC%E3%83%96%E3%83%AD%E3%82%B0:Natsugoh/%E3%83%95%E3%82%A3%E3%83%BC%E3%81%AF%E5%BF%85%E3%81%9A%E3%81%97%E3%82%82%E6%AD%A3%E8%A6%8F%E3%81%A7%E3%81%AA%E3%81%84</ref>.
 
A fee subgroup is not always normal <ref name="b">https://googology.fandom.com/ja/wiki/%E3%83%A6%E3%83%BC%E3%82%B6%E3%83%BC%E3%83%96%E3%83%AD%E3%82%B0:Natsugoh/%E3%83%95%E3%82%A3%E3%83%BC%E3%81%AF%E5%BF%85%E3%81%9A%E3%81%97%E3%82%82%E6%AD%A3%E8%A6%8F%E3%81%A7%E3%81%AA%E3%81%84</ref>.
  
 
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Revision as of 11:44, 20 September 2025

A fee subgroup \(H\) of a linearly ordered subgroup \(G\)is a subgroup such that \(g^{-1}Hg \subset H\) for all \(g > 1\).

When \(a(g) = a(h)\) implies \(a(fg) = a(fh)\) and \(a(gf) = a(hf)\) for all \(f > 1\), \(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup[1].

A fee subgroup is not always normal [2].