Abelian subgroup equals centralizer of derived subgroup in generalized dihedral group unless it is a 2-group of exponent at most four

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Statement

Suppose H is an abelian group and G is the corresponding Generalized dihedral group (?) for H. Thus, G contains H as an abelian normal subgroup of index two. Then, unless H is a 2-group of exponent at most four, H is the Centralizer of commutator subgroup (?) in G, i.e., H = C_G([G,G]).

Facts used

  1. Abelian subgroup is contained in centralizer of commutator subgroup in generalized dihedral group

Proof

Given: Generalized dihedral group G for abelian group H.

To prove: Either H is a 2-group of exponent at most four, or H = C_G([G,G]).

By fact (1), we know that H \le C_G([G,G]). Since H is a maximal subgroup of G, either H = C_G([G,G]) or G = C_G([G,G]).

In the latter case, [G,G] is in the center of G. In particular, [H,x] is in the center of G. [H,x] is the set of squares in H. For this to be in the center, we need [[H,x],x] to be trivial. But [[H,x],x] is the set of squares of elements in [H,x], which is the set of fourth powers of elements in H. Thus, the exponent of H divides 4, so H is a 2-group of exponent at most four.