# Dihedralize Twice - dihedralize a dihedral group $D_n$

A simplest dihedral group $$D_4=C_2 \ltimes C_4$$ can be regarded as dihedralizing a $C_4$ by a semi-direct product.

Q: Can one dihedralize the group $D_4$ a second time by defining $$C_2 \ltimes D_4$$ a dihedralizing a dihedral group? How to sensibly define it?

If so, what is this output nonAbelian group? (surely with 16 group elements.) Thanks. :o)

• I think you need to define "dihedralize". By the way, what has the question got to do with representation theory? – Derek Holt Apr 22 '14 at 21:51
• let us say, like this way? – annie heart Apr 22 '14 at 21:57
• As the link already says, $g\mapsto g^{-1}$ is only an automorphism if $G$ is abelian. So you have not defined "dihedralise" in the context in which you are using it. And I agree with Derek that the question has got nothing to do with representation theory, so I retagged it. – Alex B. Apr 22 '14 at 22:08

To define a semidirect product $C_2\ltimes G$ where $C_2$ acts on $G$ by inversion, you need the inversion $g\mapsto g^{-1}$ to be an automorphism on $G$. This is the case if and only if $G$ is abelian. Since $D_4=C_2\ltimes C_4$ is not abelian, you can't use this construction to dihedralize again.

If $G$ is abelian, we have $$(gh)^{-1} = h^{-1} g^{-1} = g^{-1} h^{-1},$$ so inversion is an automorphism. If conversely inversion is an automorphism, we have $$gh = ((gh)^{-1})^{-1} = (g^{-1} h^{-1})^{-1} = hg,$$ so $G$ is abelian.

• The inversion map is also an automorphism on the quaternion group (for example). – Chris Godsil Apr 22 '14 at 22:10
• Thanks both of your Chris. – annie heart Apr 22 '14 at 22:12
• @ChrisGodsil No, $(ij)^{-1} = j^{-1} i^{-1} = (-j)(-i) = ji = -k$ while $i^{-1} j^{-1} = (-i)(-j) = ij = k \neq -k$. – Christoph Apr 22 '14 at 22:16
• @Christoph: You're right. What I should have said is that generalized dicyclic groups (cs.uleth.ca/~morris/Research/Dicyclic131002.pdf) admit an automorphism that fixes or inverts each element, and so they could be "dihedralized". – Chris Godsil Apr 22 '14 at 22:23

People have already said why this doesn't have a perfect answer. I'd like to show how it doesn't even have a not-so-good answer. I'll only really address the order 16 question, so that I can be very specific. The issue is what do we mean by dihedralize. We want it to have many properties of dihedral groups, but not all, since $D_4$ does not have exactly the same properties as $C_4$ or $C_8$.

First, $D( D_4)$ should have order $2|D_4|=16$ and have $D_4$ as a subgroup. There are four solutions to this:

• the standard dihedral group
• the quasidihedral group
• the direct product, $C_2 \times D_4$
• a nonabelian rank-3 group of the form $C_2 \ltimes D_4$

However, we probably also want a semidirect product, not just a subgroup. Fortunately, $D_4$ is automatically normal (being index 2), but we need to remove the quasi-dihedral group since it is not a semi-direct product. We probably also don't want the direct product, as it silly; we would not consider $C_2 \times C_4$ to be a proper dihedral group, so $C_2 \times D_4$ seems ill-advised as $D(D_4)$.

To be really "dihedral", it seems to me that every element of $D(D_4)$ should act as either inversion or identity on each element of $D_4$: for every $x \in D(D_4)$ and $y \in D_4$, we should have $x^{-1} y x \in \{ x, x^{-1} \}$. Unfortunately none of the groups satisfy this, so I would claim there really is no sensible definition of $D(D_4)$.

However, using the "what does not already have a name needs a name" obsession, we could call $C_2 \ltimes D_4 = D(D_4)$. It is not much like a regular dihedral group, and I don't think it is very important (the “good” semidirect is $D_8$, this one is just the "not direct, but inner, semidirect product").

Also this classification becomes more complicated (and less useful) for $D_8$, where lots of the semidirect products don't have names.