Questions: Automorphism Groups and Their Structure

Question 1

In a sufficiently homogeneous model M, elements a and b realize the same complete 1-type if and only if:

• A. They satisfy the same atomic formulas that contain no parameters
• B. There exists an automorphism σ ∈ Aut(M) with σ(a) = b
• C. They have the same cardinality of definable sets containing them
• D. Their types are isolated points in the Stone space of complete types

Answer: (B)

The orbit-type correspondence says exactly this: two elements are in the same Aut(M)-orbit if and only if they realize the same complete type (in a homogeneous model). An automorphism preserves all formulas, so if σ(a) = b, then a and b satisfy exactly the same formulas — they are logically indistinguishable. The other options describe related but distinct notions that do not capture the orbit-type equivalence.

Question 2

A model M has Aut(M) = {id} — only the identity automorphism exists. What does this imply about the types realized in M?

• A. M is a minimal model with no proper elementary substructure
• B. Every two distinct elements of M realize different complete 1-types
• C. M is ω-categorical, with essentially one countable realization up to isomorphism
• D. The theory of M is complete and ω-stable

Answer: (B)

Such a model is called rigid. Since no non-identity automorphism exists, no two distinct elements can be mapped to each other — by the orbit-type correspondence, they must realize different complete 1-types. Each element is individually distinguished by some formula. This is the polar opposite of a highly homogeneous model; it indicates the theory 'pins down' individual elements precisely.

Question 3

If Aut(M) acts transitively on the elements of M (most element can be mapped to nearly every other), then different elements realize different complete 1-types.

• T. True
• F. False

Answer: False

This is exactly backwards. Transitivity of the action means every pair of elements lies in the same orbit — which by the orbit-type correspondence means they ALL realize the SAME complete 1-type. A transitive action signals maximal indistinguishability among elements, not maximal distinctness. A rich automorphism group corresponds to few distinct types, not many.

Question 4

A model in which every individual element is the unique solution to some parameter-free formula can only have the trivial automorphism (the identity).

• T. True
• F. False

Answer: True

An automorphism must preserve all formulas and their satisfaction. If element a is the unique element satisfying some parameter-free formula φ(x), then any automorphism σ must satisfy: M ⊨ φ(a) iff M ⊨ φ(σ(a)), which forces σ(a) = a. When every element is individually 'named' by a formula, no non-trivial permutation can preserve structure — the group collapses to {id}.

Question 5

Why does studying Aut(M) tell you about the complete types realized in M? Explain the key correspondence in your own words.

Model answer: An automorphism is a structure-preserving bijection, so if σ(a) = b, then a and b satisfy exactly the same formulas — they are logically indistinguishable from inside the model. Conversely, in a homogeneous model, if a and b have the same complete type (satisfy all the same formulas), you can build an automorphism mapping one to the other. So the orbits of Aut(M) acting on elements partition the elements by their complete types: same orbit ↔ same type. A large automorphism group means many elements share types (high symmetry); a small one means each element has its own type (the model is rigid).

The key insight is that automorphisms are exactly the logical symmetries of M — they cannot distinguish elements that the model's formulas cannot distinguish. This makes the group action a perfect algebraic surrogate for the logical notion of type-equivalence, allowing tools from group theory to be applied to questions about definability and types.