# Half or more elements order two implies generalized dihedral?

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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It should be true. The property is preserved under homomorphisms. Then take the minimal non-trivial normal subgroup (which is a direct product of finite simple groups). The factor-group should be either elementary abelian or has an abelian subgroup of index 2, etc. I think it should be just a standard routine proof. –  Mark Sapir Sep 26 '10 at 15:31
@Tommasso: The Z/2Z can be pulled inside the abelian part, so a direct product of a generalized dihedral group with a Z/2Z is also generalized dihedral. –  Vipul Naik Sep 26 '10 at 16:24
@Scott: I know. I just wrote the elementary abelian case separately because it is somewhat different, and also to avoid the "here's an obvious counterexample" comments/answers. But I'll clarify this in the question text. –  Vipul Naik Sep 26 '10 at 16:25
@Vipul: you are right, sorry. I guess I was missing "generalized" in your conclusion –  Tommaso Centeleghe Sep 26 '10 at 16:50
$D_8\times D_8$ ? –  Steve D Sep 26 '10 at 17:30

$D_8\times D_8$ is such a group (it has 35 involutions) that fails to be either elementary abelian or generalized dihedral.
[Reposted comment from above] @Steve: Interesting. If I understand correctly the classification you point to, then $D_8 \times D_8$ is essentially the <em>only</em> counterexample (up to padding by an elementary abelian 2-group). The counterexamples seems to stem from the properties of small numbers (namely, that (3/4)^2 > 1/2) more than any intrinsic reason. –  Vipul Naik Sep 26 '10 at 18:00
As you point out, in this comment, I missed the central product of two $D_8$s, which forms another class of examples. But these are not the only counterexamples to my conjecture -- there are a few others. There's a group of order 32 (GAP ID: 27) obtained as a semidirect product of an elementary abelian group of order 8 and an elementary abelian group of order 4 -- see groupprops.subwiki.org/wiki/SmallGroup(32,27) for a description. –  Vipul Naik Sep 26 '10 at 23:32
This is of the type "elementary abelian being acted on by element of order 2", specifically it is $(C_2)^4\rtimes C_2$. –  Steve D Sep 27 '10 at 1:22