CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-11

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The action of $Z / 2$ on two disjoint two-point orbits is free but not transitive

Statement refuted

False claim. Every free group action is transitive.

Facts & Assumptions

Given: The additive group $A = Z / 2$ and the set $X = A \times {0, 1}$ , with $a \cdot (b, i) = (a + b, i)$ .

[L1]

A left action satisfies the identity and composition laws and is transitive when one orbit is the whole set (Left group actions, transitive actions, and faithful actions).

[L2]

An action is free when $a \cdot x = x$ implies that $a$ is the identity (A free group action has no nonidentity element fixing a point).

[L3]

The orbit of $x$ is the set of all $a \cdot x$ (The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action).

[L4]

The residue classes modulo $2$ form an additive group (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L5]

The classes $0$ and $1$ are the two elements of $Z / 2$ (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ ).

Counterexample

technique · direct

1.1

The identity class satisfies $0 \cdot (b, i) = (b, i)$ , and $(a + c) \cdot (b, i) = (a + c + b, i) = a \cdot (c \cdot (b, i))$ , so [L1] and [L4] give an action.

L1L4

2.1

If $a \cdot (b, i) = (b, i)$ , then $a + b = b$ in $Z / 2$ and cancellation gives $a = 0$ ; hence the action is free by [L2].

step 1.1L2L4L5

3.1

The second coordinate is unchanged by the action, so the two sets $A \times {0}$ and $A \times {1}$ are distinct orbits by [L3]. The action is not transitive, refuting the claim.

step 1.1L1L3∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

T. W. Judson, Abstract Algebra: Theory and Applications, 14.1 (standard reference, not scraped)

The action of Z/2Z/2 on two disjoint two-point orbits is free but not transitive