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Talk:Generalized dihedral group

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Assessment comment

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The comment(s) below were originally left at Talk:Generalized dihedral group/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

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

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.

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Proof: The necessity is easily seen. Sufficiency: for every σH and τG\H, we have τσG\H, so (τσ)2 = e, or τστ-1 = σ-1. Since conjugation is an automorphism, is an automorphism of H, so H is abelian.

If [G:H] > 2, which means that there exists τ1, τ2G\H that belong to different cosets of H, then τ1τ2G\H, so for every σH 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 τG\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)[reply]