0
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ I think you need to define "dihedralize". By the way, what has the question got to do with representation theory? $\endgroup$ – Derek Holt Apr 22 '14 at 21:51
  • $\begingroup$ let us say, like this way? $\endgroup$ – annie heart Apr 22 '14 at 21:57
  • $\begingroup$ 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. $\endgroup$ – Alex B. Apr 22 '14 at 22:08
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The inversion map is also an automorphism on the quaternion group (for example). $\endgroup$ – Chris Godsil Apr 22 '14 at 22:10
  • $\begingroup$ Thanks both of your Chris. $\endgroup$ – annie heart Apr 22 '14 at 22:12
  • $\begingroup$ @ChrisGodsil No, $(ij)^{-1} = j^{-1} i^{-1} = (-j)(-i) = ji = -k$ while $i^{-1} j^{-1} = (-i)(-j) = ij = k \neq -k$. $\endgroup$ – Christoph Apr 22 '14 at 22:16
  • $\begingroup$ @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". $\endgroup$ – Chris Godsil Apr 22 '14 at 22:23
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.