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 \(G\) is linearly ordered, and \(a(g) = a(h)\) implies \(a(fg) = a(fh)\) and \(a(gf) = a(hf)\) for all \(f > 1\),
+
When \(G\) is linearly ordered and \(a(g) = a(h)\) implies \(a(fg) = a(fh)\),
or when \(G\) is bi-ordered, and \(a(g) = a(h)\) implies \(a(fg) = a(fh)\) for all \(f > 1\),
 
 
\(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup<ref name="a">https://tetrationforum.org/showthread.php?tid=1812&pid=12309#pid12309</ref>.
 
\(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup<ref name="a">https://tetrationforum.org/showthread.php?tid=1812&pid=12309#pid12309</ref>.
  

Revision as of 14:47, 8 November 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 linearly ordered and \(a(g) = a(h)\) implies \(a(fg) = a(fh)\), \(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup[1].

A fee subgroup is not always normal [2].