The question I am going to ask is really to satisfy my curiosity, as I am not at all an expert of the subject and do not plan to really work on it. Hence, if you think the question is not suitable for MO, I'll delete it. I did some browsing but could not locate any answer, but maybe I missed something simple.
Take ZF as a basis. If $Z\subset\omega$, denote by $AC(Z)$ the axiom of choice for sets of (finite) sets whose cardinality is in $Z$:
$AC(Z)$: If $x\not=\emptyset$ and for all $y\in x$ we have $|y|\in Z$, then there is a choice function on $x$.
Thus $AC(\omega)$ is the axiom of choice for sets of finite sets, and $AC(\{n\})$ (for $n\in\omega$) the axiom of choice for sets of $n$-elements sets. The implications between $AC(Z)$ for finite $Z$ and $AC(\{n\})$ have apparently been well studied in the first years of the 70s (and before by Tarski and Mostowski), I don't know to which point the question is entirely solved.
Just to give an idea, some (classical ?) implications I found in the literature are:
\begin{align*}AC(\{2\})&\Leftrightarrow AC(\{4\}),\\
AC(\{3,7\})&\Rightarrow AC(\{9\}),\\
AC(\{2,3,7\})&\Rightarrow AC(\{14\}).
\end{align*}
It is easy to see that $AC(\{n\cdot m\})\Rightarrow AC(\{n\})$:
Given $x$ containing $n$-element sets, there is a choice function on $\{y\times m\,:\,y\in x\}$ by $AC(\{n\cdot m\})$, the projection on the first factor gives a choice function on $x$. Thus, if $Z$ contains all the multiples of some given $n$, then $AC(Z)\Leftrightarrow AC(\omega)$.
My question is whether there is some kind of converse:
If $AC(Z) \Leftrightarrow AC(\omega)$, can we say something about $Z$
?
For instance, I am sure that $Z$ cannot be finite, but can it be Dedekind finite ? Must it be "big" in any appropriate sense ?
Best Answer
I think such a set must contain a multiple of $p$ for each prime $p$. Otherwise, there is nothing to rule out the possibility of a bunch of sets of size $p$ where you are able to choose an action of the cyclic group of order $p$ on each set by. no further structure. I bet you can prove this with symmetric models.
On the other hand, if you have a multiple of $p$ you get $p$, and if you have all $p\leq n$ you get $n$ by one of the classical results of this theory (I think Tarski?).
The reason this works is because of groups. Given a set of sets of size $n$, you get a set of $S_n$-torsors - the sets of total orderings of each of the sets. If, for any set of $G$-torsors, you can reduce it to a disjoint union of sets of $H$-torsors for $H$ a subgroup of $G$, by induction you can trivialize any set of $G$-torsors for any group $G$.
This induction step follows from the axiom of choice for sets of size $p$ for each prime $p$: Because every group $G$ has a subgroup of order $p$ for some $p$, you can choose one element from each orbit of that order $p$ subgroup in the set. Having done that, the symmetry group of your torsor must be reduced to some smaller group, the stabilizer of the set of chosen elements, and you can choose in some algorithmic way one coset of that stabilizer, as desired.
So the condition is: $Z$ must contain some multiple of each prime.