CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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The reflection complement in $C_{3} ⋊ C_{2} ≅ S_{3}$ is not normal

Statement refuted

A complement to the normal factor in a semidirect product must itself be normal.

Facts & Assumptions

Given: $D_{3} = C_{3} ⋊ C_{2} = ⟨ r, s : r^{3} = s^{2} = 1, s r s^{- 1} = r^{- 1} ⟩$ .

[L1]

For $n \geq 1$ the dihedral group $D_{n}$ is $Dih (C_{n}) = C_{n} ⋊ C_{2}$ of order $2 n$ , and every element has a unique form $r^{i}$ or $r^{i} s$ with $0 \leq i < n$ ; at $n = 3$ this is the dihedral group of order six ( $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations).

[L2]

The canonical complement is normal exactly when the defining action is trivial (The canonical semidirect decomposition is an internal direct product if and only if the defining action is trivial).

Counterexample

technique · direct

1.1

The subgroup $H = ⟨ s ⟩$ is the canonical complement to $⟨ r ⟩$ . The inversion action on $C_{3}$ is nontrivial because $r^{- 1} = r^{2} \neq r$ .

L1algebra

2.1

Therefore $H$ is not normal by [L2]. Explicitly, $r s r^{- 1} = r^{2} s$ , which is not in ${1, s}$ by the uniqueness in [L1].

step 1.1L1L2algebra∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 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

Keith Conrad, Semidirect Products (standard reference, not scraped)

The reflection complement in C3⋊C2≅S3C3​⋊C2​≅S3​ is not normal