I'm reading about the basic Fraenkel permutation model in Jech's The Axiom of Choice, Section 4.3.

We are working in ZF + Atoms + Axiom of Choice. $A$ is the set of atoms, assumed here to be countably infinite. $\mathscr{G}$ is the group of all permutations on $A$. Any permutation $\pi \in \mathscr{G}$ can be applied to any set $x$ in the natural way (by permuting the atoms with $x$ and its subsets, etc). We let $\operatorname{fix}(x) = \{ \pi \in \mathscr{G} : \pi y = y \text{ for all } y \in x\}$, which is a subgroup of $\mathscr{G}$ (it is the intersection of the stabilizer subgroups $\operatorname{sym}(y)$ over all $y \in x$).

Now Jech writes:

It is easy to see that the subgroup $\operatorname{fix}(A)$ is not in the filter generated by $\{\operatorname{fix}(E) : E \subseteq A \text{ finite}\}$: For every finite $E \subset A$, one can easily find $\pi \in \mathscr{G}$ such that $\pi \in \operatorname{fix}(E)$ and $\pi \notin \operatorname{fix}(A)$.

This seems clear, but also trivial. Indeed, isn't $\operatorname{fix}(A)$ just the trivial subgroup? For if $\pi a = a$ for every $a \in A$, then $\pi$ is the identity permutation. So this is just saying that for every finite set $E \subset A$, there is a nontrivial permutation that fixes $E$.

(We also have to verify that this filter $\mathscr{F}$ is a nontrivial normal filter, but that is not hard: indeed, I checked that a subgroup $H \subset \mathscr{G}$ is in $\mathscr{F}$ iff there is a finite set $E \subset A$ with $\operatorname{fix}(E) \subset H$. And since every subgroup $\operatorname{fix}(E)$ is nontrivial as mentioned, the trivial subgroup is certainly not in $\mathscr{F}$.)

I just want to make sure I am understanding this correctly, as it seems a bit odd that Jech would write out $\operatorname{fix}(A)$ without mentioning that it is really just the trivial group, if that is indeed the case.


1 Answer 1


I don't think that I've ever mentioned that $\operatorname{fix}$ anything is the trivial group in any of my works (although I may be wrong).

While it is indeed the trivial group, the important bit is that it is not in $\scr F$.

  • $\begingroup$ I should point out that in my work on iterations of symmetric extensions (the forcing-based approach for choice-related proofs) we do have natural situations where allowing the trivial group to be in the filter of subgroups is in fact useful. $\endgroup$
    – Asaf Karagila
    Mar 15, 2020 at 16:26

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