Abelian subgroup equals centralizer of derived subgroup in generalized dihedral group unless it is a 2-group of exponent at most four
Suppose is an abelian group and is the corresponding Generalized dihedral group (?) for . Thus, contains as an abelian normal subgroup of index two. Then, unless is a -group of exponent at most four, is the Centralizer of commutator subgroup (?) in , i.e., .
Given: Generalized dihedral group for abelian group .
To prove: Either is a -group of exponent at most four, or .
By fact (1), we know that . Since is a maximal subgroup of , either or .
In the latter case, is in the center of . In particular, is in the center of . is the set of squares in . For this to be in the center, we need to be trivial. But is the set of squares of elements in , which is the set of fourth powers of elements in . Thus, the exponent of divides , so is a -group of exponent at most four.