Talk:Generalized dihedral group

Assessment comment

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Moved from dihedral group. Seems like good content, but what a mess! Jim (talk) 02:56, 26 October 2010 (UTC)

Last edited at 02:56, 26 October 2010 (UTC). Substituted at 02:09, 5 May 2016 (UTC)

G is isomorphic to the generalized dihedral group of some abelian group if and only if it has a proper subgroup H such that every element in G\H has order 2.

Proof: The necessity is easily seen. Sufficiency: for every σ ∈ H and τ ∈ G\H, we have τσ ∈ G\H, so (τσ)² = e, or τστ^-1 = σ^-1. Since conjugation is an automorphism, $\sigma \mapsto \sigma ^{-1}$ is an automorphism of H, so H is abelian.

If [G:H] > 2, which means that there exists τ₁, τ₂ ∈ G\H that belong to different cosets of H, then τ₁τ₂ ∈ G\H, so for every σ ∈ H we have σ^-1 = (τ₁τ₂)σ(τ₁τ₂)^-1 = τ₁(τ₂στ₂^-1)τ₁^-1 = τ₁σ^-1τ₁^-1 = σ, which means that every element in G other than the identity has order 2, so G can be viewed as a nontrivial vector space over Z₂, which is isomorphic to the generalized dihedral group of a subspace of codimension 1. If [G:H] = 2, then for every τ ∈ G\H, we have a homomorphism {e, τ} → Aut(H), so G is the internal semidirect product of H and {e, τ}. 129.104.241.214 (talk) 15:29, 9 December 2023 (UTC)