The "generalized dihedral group" for an abelian group *A* is the semidirect product of *A* and a cyclic group of order two acting via the inverse map on *A*. *A* thus has index two in the whole group and all elements outside *A* have order two. Thus, at least half the elements of any generalized dihedral group have order two.

My question is the converse: if half or more the elements of a finite group *G* have order two, is it necessary that *G* is either an elementary abelian 2-group or a generalized dihedral group? [Note: Actually nontrivial elementary abelian 2-groups are also generalized dihedral, they're an extreme case. Also, note that the direct product of a generalized dihedral group with an elementary abelian 2-group is still generalized dihedral, because the elementary abelian 2-group can be pulled inside the abelian part.]

I have a proof when the order of *G* is twice an odd number *m*. In that case, there is a short an elegant elementary proof -- we consider the set of elements that can be written as a product of two elements of order two and show that this is a subgroup of order *m*, then show that any element of order two acts by the inverse map on it, and hence the subgroup is abelian. It can also be thought of as a toy example of Frobenius' theorem on Frobenius subgroups and complements (though we don't need Frobenius' theorem).

However, I am having some difficulty generalizing this, mainly because there are elements of order two that are inside the abelian group.

Although I have a number of possible proof pathways, I'll refrain from mentioning them for now because the actual proof is likely to be much simpler and I don't want to bias others trying the problem.

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