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Let a finite group $G$ of order $n$ be given, so $G$ is isomorphic to a permutation group embedded in $S_n$. Can we always find a set $\Omega$ such that $G$ acts transitively on $\Omega$? (For example, does $G$ always act transitively on a set $\Omega$ of size $n$)? If so, is it a difficult problem to find the smallest such set?

I am learning about Cayley's theorem and wondering about this question. I know that in general, if we're given a finite group $G$ it's hard to find the symmetric group of minimal order into which $G$ embeds. This question seems analogous in some ways, and I'm curious if it is easier.

azimut
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utx7563yu
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  • $G$ acts transitively on the orbit of any point. If, say, $G$ acts trivially on the entire set, then your $\Omega$ will have only one element. – lulu Feb 24 '24 at 17:30
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    $G$ acts transitively on itself by left multiplication. This is the essence of Cayley's theorem. – Brauer Suzuki Feb 24 '24 at 18:50
  • Ok. Can $G$ ever act transitively on a set smaller than $|G|$? – utx7563yu Feb 24 '24 at 19:41
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    to generalize the other comments $G$ acts transitively on the cosets of any given subgroup. If that subgroup is ${e}$, you get Cayley's theorem. If that subgroup is $G$, you get the trivial action. – GhostNoticer14 Feb 24 '24 at 19:42

2 Answers2

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I like to add that it is a much more interesting question to ask for a minimal set $\Omega$ such that $G$ acts faithfully on $\Omega$. Answers:

Brauer Suzuki
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  • Is any permutation group that $G$ embeds in a permutation representation of $G$? – utx7563yu Feb 26 '24 at 14:05
  • I think permutation representations are slightly more general than what I have in mind because we only need a homomorphism to a symmetric group. I am thinking about sets that a permutation group isomorphic to $G$ can act on. – utx7563yu Feb 26 '24 at 14:06
  • No. Faithful means that $G$ is isomorphic to a subgroup of the symmetric group on $\Omega$. Your original question is more general, but not very interesting, since every group acts trivially on any set. – Brauer Suzuki Feb 26 '24 at 15:41
  • I don't see how this answers the question. (I agree that regular actions are more interesting than transitive actions for this kind of question, but really this should have been a comment...) – azimut Feb 26 '24 at 20:20
  • @azimut From the OP's second comment to this answer, it is not clear what (s)he really had in mind. – Brauer Suzuki Feb 27 '24 at 06:10
  • @BrauerSuzuki The OP can ask a follow-up question of course, but as a new quesiton. The (successful) system of stackexchange is to keep the focus on the stated question within a single question-answers thread. – azimut Feb 27 '24 at 08:00
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Your question has two parts.

We fix a finite group $G$ of order $n$.

(1) Does $G$ act transitively on a set $\Omega$ of size $n$?

The answer is yes, and the reason is already found in the standard proof of Cayley's theorem, which considers the the action of $G$ on itself (so $\Omega = G$) via $$g \bullet x = gx.$$ This action is easily shown to be free and transitive (which is called regular).

To check the transitivity, let $x,y\in\Omega = G$. Then for $g = y x^{-1}$ we have $$g \bullet x = (y x^{-1}) x = y.$$

(2) Smallest size of a set $\Omega$ such that $G$ acts transitively on $\Omega$

The smallest set $\Omega$ such that there is a transitive action of $G$ on $\Omega$ is the empty set $\Omega = \emptyset$. In this case, the transitivity condition $$\forall x,y\in X : \exists g\in G : g\bullet x = y$$ is vacuously true. (Admittedly this is a bit on the esoteric side, but everything is well-defined and it is the correct answer to your question.)

Addition: General characterization of transitive actions of $G$

As requested in the comments below, we investigate non-empty sets $X$ for a transitive action of $G$ on $X$. The general answer is the following statement.

Let $n$ be a positive integer.

There is a transitive action of $G$ on a set of size $n$ if and only if $G$ has a subgroup of index $n$.

Proof

For the direction $\Rightarrow$, assume that $G$ acts transitively on a set $X$ of size $n$. let $x\in X$. By the orbit-stabilizer theorem, the stabilizer $G_x$ of $x$ has index $[G : G_x] = \#(Gx) = \#X = n$ in $G$, where the equality $\#(Gx) = n$ follows from the transitivity of the action.

For the direction $\Leftarrow$, assume that $G$ has a subgroup $H$ of index $[G : H] = n$. Then the action of $G$ on the set $G/H$ of the left cosets of $H$ via $g\bullet hH = (gh) H$ is a transitive action of $G$ on a set of size $[G : H] = n$.

Example

The alternating group $A_4$ of order $12$ has subgroups of oder $1, 2, 3, 4$ and $12$ (but remarkably not of order $6$, see this question). Hence there exists a transitive action of $A_4$ on a given set $X$ if and only if $X$ is of size $0$, $1$, $3$, $4$, $6$, or $12$.

azimut
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  • Okay, Thank you. What about if we (as we often do) ask about the smallest non-empty set $\Omega$ such that $G$ acts transitively? Is there a criterion for this in general? – utx7563yu Feb 26 '24 at 14:01
  • @utx7563yu The smallest such non-empty set has always size $1$ (look at $X = {x}$ and $g \bullet x = x$ for all $g\in G$. We can characterize the sizes $#X$ in general such that $G$ has a transitive action on $X$. I've added that to my answer. – azimut Feb 26 '24 at 20:17
  • If my answer was helpful, please don't forget to upvote... – azimut Feb 26 '24 at 20:17