I'm reading the paper "Weak forms of elimination of imaginaries" by Casanovas and Farré, where it is considered the notion of normal extension for small subsets of a big saturated model $M$:
Definition: If $A\subseteq B\subseteq M$, then $B$ is normal over $A$, or, equivalently, $B/A$ is normal, if one of the following equivalent conditions holds:
- $\forall f \in\mathrm{Aut}(M/A)$, $f(B)=B$;
- $\mathrm{Gal}(B/A)=${$f\restriction_B\,:f\in\mathrm{Aut}(M/A)$}.
Where, $\mathrm{Aut}(M/A)$ denotes the groups of all automorphisms of $M$ which fix $A$ pointwise, and $\mathrm{Gal}(B/A)$ denotes the group of all elementary permutations of $B$ which fix $A$ pointwise, respectively.
It is then observed that if $B$ is normal over $A$, then $B\subseteq\mathrm{acl}(A)$ and $\mathrm{dcl}(B)/A$ is also normal. Clearly, $B/A$ normal means that any automorphism that fixes $A$ pointwise can only permute the elements of $B$, but it is not clear to me why every element of $B$ should have a finite orbit over $A$ since $B$ could be infinite.
Is there some detail I'm missing?
Yes, $B$ could be infinite, but can't be large, by hypothesis. If there is some $b\in B$ such that $p(y)=\mathrm{tp}(b/A)$ is non-algebraic, then by saturation of $M$, $p$ has a large number of realizations in $M$. By homogeneity of $M$, the set of realizations of $p$ is the orbit of $b$ under $\mathrm{Aut}(M/A)$, so if $B$ is a normal extension of $A$, then this set is contained in $B$, and hence $B$ is large.
In the comments, you asked for clarification on why a non-algebraic type $p$ must have a large number of realizations in $M$. The basic idea is that using the saturation of $M$, we can "keep realizing $p$" until we have a large number of realizations. This idea leads naturally to a proof by transfinite induction, which works just fine. But here's a slightly more elegant "one-step" approach.
First, recall that there is a cardinal $\kappa$ such that the big model $M$ is $\kappa$-saturated and $\kappa$-strongly homogeneous. (If you assume the big model is fully saturated, then $\kappa = |M|$ and strong homogeneity follows from saturation.) Then "small" means cardinality $<\kappa$ and "large" means cardinality $\geq \kappa$.
Suppose $p(x)$ is a non-algebraic type over the small set $A$. I claim that $p(x)$ has a large number of realizations in $M$.
Since $p(x)$ is non-algebraic, there is some model $N$ containing $A$ such that $p(x)$ has infinitely many realizations in $N$. By Löwenheim-Skolem, and since $A$ is small, we can assume that $N$ is small, and thus embeds elementarily in $M$ over $A$. It follows that $p(x)$ has infinitely many realizations in $M$.
Now let $C = p(M) = \{c\in M\mid M\models p(c)\}$. Assume for contradiction that $C$ is small, and consider the type $p(x)\cup \{x\neq c\mid c\in C\}$. This is a type over the small set of parameters $A\cup C$, which is consistent by compactness: Since $p(M)$ is infinite, we can always find a realization of $p$ in $M$ which is distinct from any finitely many elements of $C$. But by definition of $C$, this type has no realizations in $M$, contradicting saturation.