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

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.