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It is said in my textbook and many proofs online that transitive abelian permutation groups are regular. However, the proofs always rely on the statement that if $G$ is a permutation group acting on $X$ and $gx = x$ for all $x,$ then $g = e.$

This is only true if the action is faithful, yet this is never mentioned. Why is this assumption constantly dropped? Permutation groups don't necessarily have to be faithful nor do the references I see assume that when defining them, you can consider $S_2$ acting on $\{1\}$ in the only possible way for example.

Sources: Projective Planes by Hughes & Piper

https://mathworld.wolfram.com/PermutationGroup.html

Dummitt & Foote

Questions and answers which make the assumption:

Proof that a transitive permutation group (G, X) with G abelian, is sharply regular

Showing that a transitive abelian permutation group is necessarily regular

Shaun
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  • Which sources are you using? In my experience, permutation groups are always assumed to have faithful actions. – Alexander Slamen Feb 14 '23 at 07:15
  • @AlexanderSlamen Updated – Display name Feb 14 '23 at 20:45
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    Group actions need not be faithful. A permutation group is a subgroup of $\mathrm{Sym}(\Omega)$ for some set $\Omega$, so the associated action is necessarily faithful. – David A. Craven Feb 14 '23 at 20:49
  • Which textbook, exactly? – Shaun Feb 14 '23 at 23:03
  • @DavidA.Craven I would agree if $\Omega = X,$ but the permutation group need not be associated with the set being acted on. – Display name Feb 15 '23 at 04:49
  • @Displayname Sure. And then it's not a permutation group, but a group action. A permutation group is defined as being a subgroup of a symmetric group. If that group acts on another set then that's not a permutation group any more, but a standard group action. – David A. Craven Feb 15 '23 at 10:04

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