CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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The action of $\mathbb Z/2$ 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 $A=\mathbb Z/2$ 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 $n$ , $(\mathbb{Z}/n,+)$ 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 $\mathbb Z/2$ (For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

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 $\mathbb Z/2$ 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

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