TITLE: Weak forms of the Axiom of Choice
QUESTION [11 upvotes]: Let $n\geq 2$ be a natural number and consider the following: 
$AC(n)$: For each family $\{X_i\}_{i \in I}$ of $n$-element sets the product $\prod_{i\in I}X_i$ is non-empty. 
Is it known that for which values of $m$ and $n$, $AC(m)$ and $AC(n)$ are equivalent !?

REPLY [3 votes]: Isn't the point that this is really all about finite groups? Given a group $G$, the $G$-axiom of choice says that for any collection of sets with a simply transitive action of $G$, we may choose from each set a coset of a proper subgroup.
Then $AC(n)$ is equivalent to the axiom of choice for all groups $G$ with a fixed-point-free action on $n$ objects.
The axiom of choice for a subgroup or a quotient group of a group implies the axiom of choice for that group.
I believe that all implications between different finite axioms of choice follow from these facts.