Difference between revisions of "Fee subgroup"
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A fee subgroup \(H\) of a linearly ordered subgroup \(G\)is a subgroup such that \(g^{-1}Hg \subset H\) for all \(g > 1\). | 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 \(f > 1\), | + | When \(G\) is bi-ordered, and \(a(g) = a(h)\) 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/wiki/User_blog:Natsugoh/Discriminating_whether_a_function_of_a_real_variable_is_an_iterated_function</ref>. | \(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) 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>. | ||
Revision as of 09:52, 21 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 \(G\) is bi-ordered, and \(a(g) = a(h)\) 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].
- ↑ https://googology.fandom.com/wiki/User_blog:Natsugoh/Discriminating_whether_a_function_of_a_real_variable_is_an_iterated_function
- ↑ 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