CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck 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

1.1L1algebra

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

2.1step 1.1L1L2algebra∎

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].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

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

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