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Motivation. I get a question, and it reads as follows: "Assume $G$ is an abelian, transitive subgroup of $S_n$" (c.f. Dummit and Foote 4.1.3). Immediately, I can tell you a ton of properties, facts, intuition, etc., revolving around group $G$ solely from the fact that it is abelian. However...notice that it is also said to be transitive. This is an important detail which likely makes the difference as to whether the group will satisfy the statement that is to be posed in the question. And yet, all I understand is (i) the definition, (ii) some basic examples (dihedral, alternating, generated by $n$-cycle), and (iii) statements (e.g. Any subgroup $<S_n$ which contains an $n$-cycle is transitive, but the converse is not true). Indeed, from this little bit of information, it holds that only the definition is truly useful.

And so my question reads as follows: What are some theorems, examples, properties, etc., which may serve beneficial in understanding the structure of transitive groups? What resources do you have? Dummit and Foote seems to delegate all introductory discussion of this topic to exercises, which is not too incredibly helpful when trying to understand the group.

I appreciate any and all help I can get. Thank you!

J.G.131
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    If there were no theorems given in your book about transitivity, then you should complete this exercise without using other theorems. I think this question would be more well-posed if you just asked directly for help with the exercise (and gave the statement of the problem in its entirety). – Malady May 21 '24 at 20:09
  • As a start - did you see the posts here at this site? For example, take this post, or this one. – Dietrich Burde May 21 '24 at 20:09
  • @DietrichBurde Yes, I have already looked into this. (Though I missed the $n\ |\ |G|$ part on my first read through.) This is a good theorem to have, though certainly is far from enough. Thanks for sharing though. – J.G.131 May 21 '24 at 20:13
  • @Malady The exercise was mentioned for demonstrative purposes. It's a pretty simple exercise, though I do realize that the result to come from it would be a good theorem to keep hold of. (We show that $|G|=|A|$ whenever $G<S_n$ is transitive and abelian.) – J.G.131 May 21 '24 at 20:15
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    Similar: https://math.stackexchange.com/questions/1196748/conditions-on-cycle-types-for-permutations-to-generate-s-n – Travis Willse May 21 '24 at 20:47
  • @TravisWillse Neat result! Thanks for sharing. So given a transitive group containing an $(n-1)$-cycle and transposition, we deduce that this group $=S_n$. Got it! – J.G.131 May 21 '24 at 21:03
  • As a start, if $G$ is a transitive subgroup of $S_n$, then $n$ divides $|G|$. – Kan't Jun 08 '24 at 04:08

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In my opinion, the most important basic fact about transitive groups is the following:

A transitive (right/left) action of a group $G$ on $\Omega$ is equivalent to the (right/left) action of $G$ on the (right/left) cosets of $G_\omega$ by (right/left) multiplication (for fixed $\omega\in\Omega$).

This is very easy to prove and is used almost without thinking by anyone who often works with these things. It provides a sort of dictionary between the language of actions and internal properties of the group. For example,

  • the kernel of the action is the core of $G_\omega$ in $G$, so
  • the action is faithful (i.e. you can think of it as a permutation group) if and only if $G_\omega$ is corefree
  • there is a one-to-one correspondence between blocks of imprimitivity of $G$ containing $\omega$ and subgroups of $G$ containing $G_\omega$, which implies that
  • $G$ acts primitively if and only if $G_\omega$ is maximal in $G$,

etc...

Note that this is a characterisation, so nothing is ever lost by translating, but sometimes it can help to see some things.

For example, let us try and apply this to your situation. You have a group $A$ acting transitively on $\Omega$. Translating, this means that we have an abelian group with a core-free subgroup $A_\omega$ (corresponding to a point-stabiliser). But in an abelian group, every subgroup is normal so only the trivial subgroup can be core-free. Therefore $A_\omega=1$ which is equivalent to $A$ being regular.

(Which is maybe what you were asked to prove? Note that we didn't really need $A$ to be a permutation group, an action was sufficient, we got faithfullness for free.)

verret
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