Symmetric Group of an amorphous set An amorphous set is an infinite set, which is not the disjoint union of two infinite sets. The existence of such sets is consistent with ZF.
I am wondering, if there are any interesting remarks to be made about the structure of $G:=\operatorname{Sym}(A)$, the symmetric group on some amorphous set $A$.
In particular, do we know something about the "cardinality" (in a ZF-sense) or a minimal generating set of $G$?
One trivial thing I was able to notice is that any bijection in G either has finite or cofinite support. And at least those of finite support allow for a generating set of size $A$, namely those permutations $(1\,\; a)$ where $1$ is some fixed element of $A$ and $a$ is arbitrary.
 A: Well. There's not a whole lot we can say at this level of generality. For example, if $A$ is strongly amorphous, i.e. every partition of $A$ is almost entirely made of singletons, then it is easy to see that any permutation of $A$ can only move finitely many points, since it decomposes $A$ into orbits, and those must be finite.
On the other hand, if $A$ is a $2$-amorphous set, i.e. there is a partition into pairs, then there is also a permutation which simply switches the elements of each pair, and therefore has infinitely many nontrivial orbits.
On the third hand,1 if $A$ is amorphous, then any infinite partition of $A$ must also be amorphous, and so it is easy see that the behaviour across the different orbits of a given permutation must be the same almost everywhere.
Finally, and this is where it starts getting really weird, there's a lot of structure that is compatible with certain "level" of amorphous-ness2 sets. Given any first-order structure whose first-order definable subsets are finite or co-finite, and has sufficiently rich automorphism group (enough to witness strong homogeneity), we can arrange a model where this structure has an amorphous "version".
For example, an infinite dimensional vector space over a finite field. But also a set with some partitions that satisfy certain basic rules can be used.
So, if we have an infinite set with a partition into $3$-element sets, we can also add a function to this structure, which is a permutation that generates that partition; but we can also elect to not add such a function, and ensure that none exist, i.e. that a $3$-amorphous set will only have finitely supported permutations.
Therefore, at this "maximal" level of generality, of just talking about amorphous sets, there's not a whole lot that we can say. Sam Tarzi, a former student of Peter Cameron, wrote a Ph.D. thesis titled "Group Actions on Amorphous Sets and Reducts of Coloured Random Graphs" and they have a preprint (presumably derived from that work, although seems to have never been through the peer review process), titled "Group Actions On Amorphous Sets" that might be of interest to you.
Here's something that I couldn't find in the literature, but I took a few minutes to make the argument. But I am almost certain this should be known.
Theorem. If $A$ is amorphous, then $\operatorname{Sym}(A)$ is Dedekind-finite.
Proof. Suppose that $\{p_n\mid n<\omega\}$ is a sequence of distinct permutations of $A$. Recall that $[A]^{<\omega}$, the set of finite subsets of $A$, is Dedekind-finite and the each orbit of $p_n$ must be finite.
For a given $a\in A$, let $O(a)$ be the closure of $\{a\}$ under all the $p_n$s. We first claim that $O(a)$ must be finite. Since orbit of $a$ under each $p_n$ must be finite, there must be finitely many "types" of orbits, as $[A]^{<\omega}$ is Dedekind-finite; taking $O_0(a)$ as the union these orbits, we define $O_{n+1}(a)=\bigcup\{O(x)\mid x\in O_n(a)\}$, then each $O_{n+1}$ is finite, and this is an increasing sequence of sets in $[A]^{<\omega}$, so it must stabilise, and therefore $O(a)$ must be finite.
Next, as $\{O(a)\mid a\in A\}$ must form a partition of $A$, all but finitely have the same size. More importantly, though, all of the interesting properties that follow from how $p_n$, for whatever $n$, acts on $O(a)$ must also be the same for all but finitely many examples.
In particular, for a fixed $n$, all but finitely many $a$s have the same "type" of $p_n$ on $O(a)$. In other words, for a cofinite set of $a$ and $b$ there is a bijection mapping $O(a)$ to $O(b)$ which commutes with $p_n$.
So, letting $A_n$ be the [finite] set of "atypical $O(a)$", we have, by Dedekind-finiteness, that $\{A_n\mid n<\omega\}$ is just finitely many sets, repeating. Therefore $A^*=\bigcup A_n$ is a finite set, and for any $a,b\notin A^*$, $p_n$ has the same type on $O(a)$ and $O(b)$. But since each $O(a)$ is finite, there can only be finitely many of these types, so must be able to distinguish infinitely many $p_n$s just by examining $A^*$.
But now we ran into a bit of a problem! Since $A^*$ is finite, there can only be finitely many permutations of $A^*$, so we cannot distinguish infinitely many permutations at once!
So, the only way to resolve this is that there was no countably sequence of permutations, and therefore $\operatorname{Sym}(A)$ is Dedekind-finite as well.     $\square$


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*Amorphous sets are weird, and apparently they have more than two hands. Don't get me started on the number of fingers each of those have.


*Amorphity? amorphousity? amorphousnessicity? who knows...
