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 \(G\) | + | When a map \(a\) defined on a linearly ordered group \(G\) satisfies that \(a(g) = a(h)\) 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:// | + | \(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup<ref name="a">https://tetrationforum.org/showthread.php?tid=1812&pid=12309#pid12309</ref>. |
| − | A fee subgroup is not always normal <ref name="b">https://googology.fandom.com | + | A fee subgroup is not always normal <ref name="b">https://googology.fandom.com/wiki/User_blog:Natsugoh/A_fee_subgroup_is_not_necessarily_normal</ref>. |
<references /> | <references /> | ||
Latest revision as of 14:50, 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 a map \(a\) defined on a linearly ordered group \(G\) satisfies that \(a(g) = a(h)\) implies \(a(fg) = a(fh)\) for all \(f > 1\), \(P(a) = \{g^{-1}h \mid a(g) = a(h)\}\) is a fee subgroup[1].
A fee subgroup is not always normal [2].