Number of automorphisms of saturated models

I have the following assignment question: Let $M$ be an $L$-model of cardinality $\kappa$. Assume $M$ is saturated. How can you show that $|\text{Aut}(M)|=2^{|M|}$?

I see two possible ideas/connections/intuitions here:

1. Definable sets. Since $M$ is saturated, these are either finite or of cardinality $\kappa$. Then maybe you can somehow use the fact that these are preserved by automorphisms?

2. Maybe some sort of diagonal argument. If you try to capture $\text{Aut}(M)$ with $\lambda<2^{|M|}$ automorphisms, then you can show that you'll be missing at least one.

There's this question, whose title was originally going to be my title, but I wanted to avoid confusion. While it doesn't really answer my question, perhaps the idea of moving non-definable points via automorphisms could yield the required cardinality for $\text{Aut}(M)$. I imagine using the finite definable sets (I believe these are called algebraic, but correct me if I'm wrong), and "permutating" the points outside of these sets? Then perhaps it becomes an easy cardinality argument...

I'm mostly thinking out loud, as I'm not sure how to make all of these ideas concrete, and I'm not even sure if they are on the right path. Help?

Solution 1:

I decided to write up an answer, so that this question can be removed form the list of unanswered questions.

First enumerate $M = \{m_\alpha : \alpha < \kappa\}$. For every $s \in 2^{<\kappa}$ we construct a partial automorphism $f_s$. I.e. $f_s : M \to M$ is an elementary map. The idea is that these automorphisms extends one another ($t \subseteq s$ implies $f_t \subseteq f_s$) but $f_{s0}$ and $f_{s1}$ disagree on some element. In the end for $q \in 2^\kappa$ we define $f_q = \bigcup_{\alpha < \kappa} f_{q|_\alpha}$, which by the above is a well defined elementary map and $f_q \neq f_p$ for $q \neq p \in 2^\kappa$. We also throw in the back and forth condition to make $f_q$-s automorphisms.

More precisely

• Let $f_\emptyset = \emptyset$.
• If $\delta$ is a limit ordinal and $s \in 2^\delta$, let $f_s = \bigcup_{\alpha < \delta} f_{s|_\alpha}$
• If $s \in 2^{\alpha+1}$ consider two cases. If $\alpha$ is an even ordinal, then pick $m \in M$ of minimal index not in the domain of $f_{s|_\alpha}$. By saturation pick $n_0, n_1$ in $M$ such that $f_{si} = f_s \cup \{\langle m, n_i \rangle\}$ are elementary. If $\alpha$ is an odd ordinal, go back in a similar way. I.e. pick $n \in M$ with the least index not in the image of $f_{s|_\alpha}$ and find two distinct element to extend $f_{s|_\alpha}$ with.