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

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$S_{3}$ acting on three points has $∣ X ∣ = 3$ and $∣ X^{S_{3}} ∣ = 0$ , so the fixed-point congruence modulo $2$ fails without the $p$ -group hypothesis

Statement refuted

False claim. For every finite group $G$ acting on a finite set $X$ , one has $∣ X ∣ \equiv ∣ X^{G} ∣ (m o d 2)$ .

Facts & Assumptions

Given: The natural action of $S_{3}$ on $X = {1, 2, 3}$ .

[L1]

The fixed-point congruence is proved for finite $p$ -groups (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ ).

[L2]

The global fixed set consists of the points fixed by every group element (The fixed-point sets $X^{g}$ and $X^{G}$ of a group action).

[L3]

The natural $S_{3}$ -action on three points is faithful and transitive (The natural action of $S_{3}$ on three points is faithful and transitive but not free).

[L4]

The group $S_{3}$ has order $6$ (The class equation of $S_{3}$ is $6 = 1 + 2 + 3$ ).

[L5]

Congruence modulo $2$ means divisibility of the difference by $2$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

Counterexample

technique · direct

1.1

For each $i \in X$ , some transposition moves $i$ , so no point is fixed by every element of $S_{3}$ and $X^{S_{3}} = \emptyset$ . Thus $∣ X ∣ = 3$ and $∣ X^{S_{3}} ∣ = 0$ .

L2L3

2.1

The difference $3 - 0 = 3$ is not divisible by $2$ , so [L5] shows that the congruence fails. By [L4], $∣ S_{3} ∣ = 6$ is not a power of $2$ , so this does not contradict [L1] and isolates its $p$ -group hypothesis.

step 1.1L1L4L5algebra∎

Depends on

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Dependency tree · two levels

20 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

K. Conrad, Group Actions, Theorem 4.1 and following discussion (standard reference, not scraped)

S3S3​ acting on three points has ∣X∣=3∣X∣=3 and ∣XS3∣=0∣XS3​∣=0, so the fixed-point congruence modulo 22 fails without the pp-group hypothesis