# Basic Fraenkel model in Jech's "The Axiom of Choice"

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.

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 $$\scr F$$.