Talk:Generalized dihedral group

Assessment comment
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)