G = ℤ₄ acts on the four vertices of a square by rotation. The orbit of vertex v₁ is {v₁, v₂, v₃, v₄} and the stabilizer of v₁ is {0}. What does the orbit-stabilizer theorem predict about |G|?
A|G| = |Orb(v₁)| + |Stab(v₁)| = 4 + 1 = 5
B|G| = |Orb(v₁)| × |Stab(v₁)| = 4 × 1 = 4
C|G| = |Stab(v₁)|² = 1, since ℤ₄ acts freely on v₁
DThe theorem does not apply here because the action is transitive
The orbit-stabilizer theorem states |Orb(x)| · |Stab(x)| = |G| for any x in X. Here |Orb(v₁)| = 4 (all four vertices are reachable from v₁) and |Stab(v₁)| = 1 (only the identity fixes v₁), so 4 × 1 = 4 = |ℤ₄|. The formula multiplies, not adds. The theorem applies to all group actions — transitivity (a single orbit covering all of X) is not required; apply it element by element.
Question 2 Multiple Choice
Which of the following correctly describes a group action of G on a set X?
AA bijection σ: G → X assigning each group element a unique point in X
BA function · : G × X → X satisfying e·x = x for all x, and (gh)·x = g·(h·x) for all g,h ∈ G and x ∈ X
CAny function G × X → X, provided G is abelian and X is finite
DA surjective group homomorphism from G onto the symmetric group Sym(X)
The definition requires two axioms: the identity axiom (e·x = x — the identity element acts trivially) and the compatibility axiom ((gh)·x = g·(h·x) — the group multiplication is compatible with the action). These axioms ensure that the map g ↦ (x ↦ g·x) is a group homomorphism G → Sym(X). Option A defines a function G → X, which is the wrong type. Option D describes an action only if the map is surjective, but actions need not induce surjections onto Sym(X).
Question 3 True / False
The axioms of a group action guarantee that each group element g induces a bijection on X — that is, the map x ↦ g·x is a permutation of X.
TTrue
FFalse
Answer: True
The map φ_g: X → X defined by φ_g(x) = g·x has a two-sided inverse: φ_{g⁻¹}(x) = g⁻¹·x. The compatibility axiom gives φ_g(φ_{g⁻¹}(x)) = g·(g⁻¹·x) = (g g⁻¹)·x = e·x = x, and symmetrically in the other order. So every φ_g is a bijection, i.e., a permutation of X. This is precisely why the map G → Sym(X) sending g to φ_g is well-defined.
Question 4 True / False
If a group G acts on a set X, there is expected to be exactly one orbit — most element of X is reachable from most other by some group element.
TTrue
FFalse
Answer: False
An action with a single orbit is called transitive, but this is a special case, not a requirement. In general, a group action partitions X into orbits, and there can be many. For example, G = ℤ₂ acting on X = {1, 2, 3, 4} by swapping 1↔2 and fixing 3 and 4 produces three orbits: {1,2}, {3}, and {4}. The orbit-stabilizer theorem applies to each orbit separately, and each orbit's size divides |G| by Lagrange's theorem.
Question 5 Short Answer
State the two axioms of a group action and explain why they together imply that a group action defines a group homomorphism from G into Sym(X).
Think about your answer, then reveal below.
Model answer: The two axioms are: (1) e·x = x for all x ∈ X (the identity acts trivially); (2) (gh)·x = g·(h·x) for all g,h ∈ G and x ∈ X (compatibility with group multiplication). To see why these define a homomorphism φ: G → Sym(X): define φ(g) to be the function x ↦ g·x. Axiom (1) ensures each φ(g) is invertible with inverse φ(g⁻¹), so φ(g) ∈ Sym(X). Axiom (2) ensures φ(gh)(x) = (gh)·x = g·(h·x) = φ(g)(φ(h)(x)), which means φ(gh) = φ(g)∘φ(h) — exactly the homomorphism condition.
This homomorphism perspective unifies all group actions: every action is secretly a way of representing G as a group of symmetries (permutations) of some set. Cayley's theorem is the special case where X = G and the action is left multiplication, showing every group embeds in its own symmetric group.