A group G of order 24 acts on a finite set. If the stabilizer of a point x has order 6, what is the size of the orbit of x?
A4
B6
C18
D24
By the orbit-stabilizer theorem, |Orb(x)| · |Stab(x)| = |G|, so |Orb(x)| = 24 / 6 = 4. The common mistake is to subtract rather than divide: |G| − |Stab(x)| = 18 is incorrect. The theorem is a multiplicative relationship, not additive, because the orbit corresponds to the cosets of the stabilizer in G.
Question 2 Multiple Choice
The group S₃ acts on itself by conjugation. The orbit of (123) has size 2 (the two 3-cycles). What is the order of its stabilizer?
A1
B2
C3
D6
|Stab((123))| = |G| / |Orb((123))| = 6 / 2 = 3. The stabilizer consists of all elements that commute with (123) under conjugation — these are the identity and the two 3-cycles themselves, forming a cyclic subgroup of order 3. This illustrates how the theorem lets you compute the stabilizer's size from orbit size alone, without listing stabilizer elements.
Question 3 True / False
The stabilizer of any point x under a group action is always a subgroup of G.
TTrue
FFalse
Answer: True
Stab(x) = {g ∈ G : g·x = x} is closed under the group operation (if g and h fix x, then (gh)·x = g·(h·x) = g·x = x), contains the identity, and is closed under inverses (if g·x = x, then x = g⁻¹·(g·x) = g⁻¹·x). These three properties make it a subgroup. This fact is what allows Lagrange's theorem to be applied in the proof of the orbit-stabilizer theorem.
Question 4 True / False
If two points x and y in a set have the same stabilizer subgroup, they is expected to lie in the same orbit.
TTrue
FFalse
Answer: False
Having equal stabilizers does not imply being in the same orbit. Consider Z₂ = {e, r} acting on a three-element set {a, b, c} where r fixes both a and b but swaps nothing (trivial action on a and b) — both a and b have Stab = G, yet they are each their own orbit. The orbit-stabilizer theorem links the *size* of the orbit to the *size* of the stabilizer, but two points with equal-sized stabilizers can still be in different orbits.
Question 5 Short Answer
Why do two group elements g and h produce the same orbit image (g·x = h·x) if and only if they belong to the same left coset of Stab(x) in G?
Think about your answer, then reveal below.
Model answer: g·x = h·x iff h⁻¹g·x = x iff h⁻¹g ∈ Stab(x) iff g and h are in the same left coset of Stab(x). So each coset of Stab(x) corresponds to exactly one element of the orbit.
This coset correspondence is the heart of the proof: distinct cosets map to distinct orbit elements, and every orbit element is hit by exactly one coset. The number of left cosets of Stab(x) in G equals |G|/|Stab(x)| by Lagrange's theorem, so |Orb(x)| = |G|/|Stab(x)|. This is why the theorem feels like an extension of Lagrange's theorem from subgroups to group actions.