CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $|X|\equiv|X^G|\pmod2$ .

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 $|X|\equiv|X^P|\pmod p$ ).

[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: $a\equiv b\pmod n$ when $n\mid(a-b)$ , 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 $X^{S_3}=\varnothing$ . 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

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: 61 results over 12 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

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

S3S_3S3​ acting on three points has ∣X∣=3|X|=3∣X∣=3 and ∣XS3∣=0|X^{S_3}|=0∣XS3​∣=0, so the fixed-point congruence modulo 222 fails without the ppp-group hypothesis