CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-11

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The subgroup $⟨ (1 2 3), (1 2) (3 4) ⟩ \leq S_{4}$ has order $12$ but no subgroup of order $6$ , so Cauchy's theorem does not extend to composite divisors

Statement refuted

False claim. If a positive integer $d$ divides the order of a finite group $G$ , then $G$ has a subgroup of order $d$ .

Facts & Assumptions

Given: In $S_{4}$ , let $a = (1 2 3)$ , $b = (1 2) (3 4)$ , and $A = ⟨ a, b ⟩$ .

[L1]

Cauchy's theorem supplies an element of order $p$ when the prime $p$ divides a finite group order (Cauchy's theorem: if a prime $p$ divides $∣ G ∣$ , then $G$ has an element of order $p$ ).

[L2]

A subgroup of index $2$ is normal (Every subgroup of index two is normal).

[L3]

The symmetric group contains all permutations under composition (The symmetric group $Sym (X)$ : the bijections of a set $X$ under composition, $Sym (X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

[L4]

The generated subgroup is the smallest subgroup containing the generators (The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, Subgroup).

[L5]

For a finite group, $∣ G ∣ = [G : H] ∣ H ∣$ (Lagrange's theorem: $∣ G ∣ = [G : H] ∣ H ∣$ for every subgroup $H$ of a finite group $G$ ).

[L6]

Conjugation by a group element is an automorphism (Conjugation $x \mapsto g x g^{- 1}$ is an automorphism).

Counterexample

technique · contradiction

1.1

Direct multiplication shows that $A$ consists of the identity, the eight $3$ -cycles on ${1, 2, 3, 4}$ , and the three double transpositions $(1 2) (3 4)$ , $(1 3) (2 4)$ , $(1 4) (2 3)$ . This set is closed under multiplication by $a^{\pm 1}$ and $b = b^{- 1}$ , and every listed element is a word in $a, b$ , so [L3] and [L4] give $∣ A ∣ = 12$ .

L3L4algebra

2.1

Suppose, for contradiction, that $H \leq A$ has order $6$ . Then [L5] gives $[A : H] = 2$ , so $H ⊴ A$ by [L2]. Since $3 ∣ ∣ H ∣$ , [L1] gives an element $c \in H$ of order $3$ , necessarily one of the eight $3$ -cycles in step 1.1.

assume-contrastep 1.1L1L2L5

3.1

The subgroup $H$ contains $c^{- 1}$ . Normality and [L6] make it contain every $A$ -conjugate of both $c$ and $c^{- 1}$ . Conjugating by the displayed generators gives two disjoint four-element classes, together all eight $3$ -cycles, so $H$ would contain at least eight elements, contradicting $∣ H ∣ = 6$ .

step 1.1step 2.1L3L6algebra

4.1

Thus $6$ divides $∣ A ∣ = 12$ but is not the order of a subgroup of $A$ . The arbitrary-divisor claim is false, whereas [L1] remains valid for prime divisors.

step 1.1step 3.1L1discharge-contradiction∎

Depends on

Used by

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Sources

K. Conrad, No Subgroup of A4 Has Index 2, Theorem 1 (standard reference, not scraped)

The subgroup ⟨(1 2 3),(1 2)(3 4)⟩≤S4⟨(123),(12)(34)⟩≤S4​ has order 1212 but no subgroup of order 66, so Cauchy's theorem does not extend to composite divisors