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I'm studying transitive and regular groups comparing the theorems and corollaries on several texts I have. One reference for sure, is Dixon's Permutation Groups. Actually, I stumbled into a corollary about transitive groups at p.8, sounding like this:

Suppose that $G$ is transitive in its action on the set $\Omega$. Then:

(i) the stabilizers $G_\alpha (\alpha\in\Omega)$ form a single conjugacy class of subgroups of $G$

(ii) The index $|G : G_\alpha| = |\Omega|$ for each $\alpha$

(iii) If $G$ is finite then the action of $G$ is regular $\iff |G|=|\Omega|$

While I ha no problems with (i) and (ii), that $\iff$ is disturbing me a bit. As far as I can it means "if and only if" made by $a\Rightarrow b$ and $a\Leftarrow b$. In our case this would read:

regularity $\Rightarrow |G|=|\Omega|$ and $|G=\Omega| \Rightarrow$ regularity

On other sources, I found a less strong implication, i.e. only: regularity $\Rightarrow |G|=|\Omega|$

I know a regular group shows $G_\alpha = {e}, \forall a\in\Omega$ so the order matching should be related to the number of cosets of the trivial stabilizers, but I don't exactly how to prove this "feeling".

May you help me on that, please? Thanks

riccardoventrella
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    It is unclear where you are stuck. If you assume that $|G| = | \Omega|$ then you can deduce that $|G_\alpha|=1$ for all $\alpha$. And from that you need to deduce that the action is regular. But is that not the definition of a reular action? – Derek Holt Oct 26 '19 at 09:30
  • The problem is exactly I'm not able to deduce $|G_\alpha| = 1$, this was the question at the end. However, my real problem is with this thesis: https://www.famnit.upr.si/files/zakljucna_dela_repo/63 If you look to the definition 2.10 "A group $G$ acting transitively on a set $\Omega$ is said to act regularly if $Stab_G(\alpha) = {e} then |G|=|\Omega|$. As you can see here is not a iff, but only a if. – riccardoventrella Oct 26 '19 at 09:43
  • OK, I probably doomed it: 1) G is transitive, so $\alpha, \beta\in\Omega, \exists g\in G | \alpha^g = \beta$ 2) $|G|=|\Omega|$ Then form 2) we know $g$ from 1) must be unique, so there's a bijection $\alpha \iff g$ We also know from orbit stabilizer theory $\beta = \alpha^{G_\alpha g}$, but since $g$ is unique $\rightarrow G_\alpha = {e}$, so the action is regular. Is this correct from your POV? – riccardoventrella Oct 26 '19 at 11:47
  • I still don't understand where you are stuck. The fact that $|G_\alpha|=1$ follows immediately from (ii), and you said that you had no problems with (i) and (ii). – Derek Holt Oct 26 '19 at 12:38
  • Have you read my comment right above yours? I'm asking if my reasoning is correct. If yes, as I guess, I'm no more stuck. However, the pdf thesis I posted above has not an iff, and so is wrong. – riccardoventrella Oct 26 '19 at 13:05
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    For definitions, it is traditional to use "if" to mean "iff". In other words, in a definition, if someone say "a blah is regular if it satisfies P", it usually mean "iff it satisfies P". – verret Oct 26 '19 at 17:48
  • I wasn't aware of that and it seems to me a but misleading, but of course a convention is a convention, so no concern on that. Actually it was exactly that if instead of iff which let me go deeper into the theory, since now I fully understood the iif in Dixon's book. Thank for pointing that out, @verret – riccardoventrella Oct 26 '19 at 18:14
  • I'll make it an answer so you can accept it. – verret Oct 26 '19 at 19:57

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For definitions, it is traditional to use "if" to mean "iff". See

Are "if" and "iff" interchangeable in definitions?

verret
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