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I came across this problem studying for a qualifier, and I'm pretty stumped outside of the more obvious steps to take. Here is what I've done so far, which isn't much at all.

Let $A_4$ act on a set $S$ transitively. Then $A_{4} \cdot x = S$ for every $x \in S$. The trivial action implies $A_4 \cdot x = \{x\}$, so one possibility is $\text{card}(S)=1$. I also know that $A_4$ has various subgroups, one normal of order $4$, several of order $2$ and of order $3$, but I'm not sure how this helps.

I also know that any set $S$ which $A_4$ acts on is nessiarily partitioned into by orbits $S = \bigcup_{x} \text{orb}(x)$ with $\sum_{x} \text{card}(\text{orb}(x)) = \sum_{x} [A_4:(A_{4})_x]$ by the orbit stabilizer theorem, where $x$ ranges over a collection of distinct orbit representatives, but I'm not sure where to go from here.

Any help would be appreciated. Thank you.

Isochron
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  • A transitive group has exactly one orbit, and this orbit must be $S$. So, $|S|\big||A_4|$. Does that help narrow things down? – Chris Jan 14 '23 at 04:41
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    You need to make the following connection: "transitive" is another way of saying "there is exactly one orbit". Make sure you understand why this is true! I think you can take it from there. – diracdeltafunk Jan 14 '23 at 04:46
  • @diracdeltafunk I see why this is true, and we have exactly that $|S|= A_4/(A_4)_x$, which give you that all possible values of $|S$ are the divisors of 12, but I’m still unsure how one proves this statement, since I don’t know if $|S|$ can actually take every one of these values, just that $|S| \mid A_4$ – Isochron Jan 14 '23 at 05:22
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    Right, you must determine the possible orders of a stabilizer. Fact: every subgroup is a possible stabilizer -- in other words, for every group $G$ and every subgroup $H \leq G$, there exists a transitive $G$-set $X$ and an element $x \in X$ such that $G_x = H$. Can you see why? – diracdeltafunk Jan 14 '23 at 05:57
  • @diracdeltafunk The closest thing I can think of is letting G act on the set of all subgroups of H by conjugation, but I think that’s N_G(H), not H. – Isochron Jan 14 '23 at 06:44
  • @diracdeltafunk So letting G act on the set of all left cosets of H by left translation does it I believe. H itself is a left coset, and G_H ={g in G | gH = H} which is equivalent to {g in G | g in H} = H. So then the answer to this question is that Every number 12/|H| is a possible cardinality for S for H a subgroup of A_4? Since given any subgroup H in A_4, there exists by above a set S which A_4 acts transitivity on, where |S| = |A_4/H|? This seems almost like the converse to what I’m looking for though, since I’m given a set S apriori which A_4 acts transitive on.. – Isochron Jan 14 '23 at 07:43

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If a finite group $G$ acts transitively on a set $X$, then there is a subgroup $H \le G$ and a $G$-equivariant bijection between the set $(G : H)$ of cosets of $H$ and $X$. Hence it can act transitively on a set $S$ of cardinality $n$ if and only if $G$ has a subgroup of index $n$.

For $G=A_4$ is is well-known, for which index $n$ there exists a subgroup. Recall that there is no subgroup of index $2$ in $A_4$, see below:

$A_4$ has no subgroup of order $6$?

So from the list of divisors of $|A_4|=12$, all except one divisor are possible cardinalities for $S$.

Dietrich Burde
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