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The subgroup $\langle(1\,2\,3),(1\,2)(3\,4)\rangle\le S_4$ 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 $A=\langle a,b\rangle$ .

[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 $\operatorname{Sym}(X)$ : the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ 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 $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , 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 $x\mapsto gxg^{-1}$ 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 $a^{\pm1}$ 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 $H\le A$ has order $6$ . Then [L5] gives $[A:H]=2$ , so $H\mathrel{\trianglelefteq}A$ by [L2]. Since $3\mid|H|$ , [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∎

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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\langle(1\,2\,3),(1\,2)(3\,4)\rangle\le S_4⟨(123),(12)(34)⟩≤S4​ has order 121212 but no subgroup of order 666, so Cauchy's theorem does not extend to composite divisors