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I want to make sure I am correct about something I read in Moore and Pollatsek's "Difference Sets":

Suppose we have an abelian group $H$. If $\exists g\notin H$ with $g^2=1$ and $ghg^{-1}=h^{-1}\;\forall h\in H$ such that a group $G$ can be written as $H+gH$ in the integral group ring $\mathbb{Z}G$, then $G$ is called a generalized dihedral extension of $H$.

The question arises, since they don't make it explicit in the definition, whether $g\in G$, necessarily. I think this is true.

It certainly can be, for they give the clear example in the text where $H=\langle a,b|a^6=b^2=1, ab=ba\rangle$ and $G=\langle a,b,c| a^6=b^2=c^2=1, ab=ba, ac=ca^{-1}, bc=cb\rangle$ is a dihedral extension of $H$ where $c$ is the extending element.

I think it is necessary that $g\in G$ because if $G=H+gh\in\mathbb{Z}G$, then the elements of $G$ are precisely those of $H$ and those of the form $gh$ with $h\in H$. Since $H$ must have an identity, there is an $h\in H$ such that $gh=g$ and so $g\in G$. I don't know why I am doubting myself, but could someone confirm? I'll sleep better.

Perhaps there is an easier way to see that $g\in G$? Or if I'm wrong, an easy counterexample?

The Count
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  • I don't follow your reasoning at all. Why do you say that the elements of $G$ are those of $H$ and those of form $gh$? A general element of ${\mathbb Z}G$ has the form $h_1 + gh_2$. But I might be misunderstanding, because you never really said what $G$ was! – Derek Holt Apr 04 '17 at 08:42
  • @DerekHolt $G=H+gH$. For some $g\notin H$ that meets the other states criteria. That's all. And my understanding of $\mathbb{Z}G$ is that a general element is $\sum_ga_gg$, from the book I'm using, mentioned above... – The Count Apr 04 '17 at 12:03
  • But then it's clear that $g \not\in G$, since $g \ne h_1 + gh_2$ for any $h_1,h_2 \in H$. – Derek Holt Apr 04 '17 at 12:38
  • The example clearly has $g\in G$, though, so I think we are not communicating well... – The Count Apr 04 '17 at 12:57
  • Try and clarify your post. For example the statement $g \not\in H$ does not make immediate sense, because you have not said what superset of $H$ $g$ is being assumed to lie in. – Derek Holt Apr 04 '17 at 16:01
  • That's how it is presented in the book. The odd mention of $g\notin H$ without specifying a superset was the entire impetus behind my asking the question, I fear. I promise I'm not trying to be difficult! – The Count Apr 04 '17 at 18:34

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This is a terribly presented definition. A much more reasonable definition would be as follows:

Let $H$ be an abelian group. A group $G$ is the GDE of $H$ if it is the external semidirect product of $H$ with $C_2$, where conjugation by the non-identity element acts as the inverse map on $H$. That is, it is the group formed by $G=H\rtimes_\phi C_2$ where $\phi_c(h)=chc^{-1}=h^{-1}$ with $c$ being the non-identity element of $C_2$.

You can see this definition being used here and here. With this rephrasing, the book's definition's mention of $\mathbb{Z}G$ doesn't make a whole lot of sense (or maybe it does, it's been a while since I've done this) but the rest of what is said does. Why they chose to write it like that escapes me.

So yes, $g\in G$. In fact, $g\in G$ is crucial to the construction of $G$ in the first place! It's the image of the non-identity element of $C_2$ when mapped into $G$ as a subgroup in this construction.

Note that the motivation for this is that if you let $H=C_n$ then you get the GDE of $H$ is $D_{2n}$, so this is an extension of a group to look like a dihedral group.

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Stella Biderman
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  • But the question that that apparently being asked related to the group ring ${\mathbb Z}G$. Was that a complete misunderstanding? – Derek Holt Apr 05 '17 at 06:57
  • @DerekHolt Ah, forgot about that part. I don't know how to make sense of that (though $G=H+gH$ works in this definition) – Stella Biderman Apr 05 '17 at 11:47
  • Yeah, I didn't like the definition either. The rest of the book so far is very nice. Quick clarification point before I go any farther: Did you mean the second sentence in your definition? Should it say "external direct product of $H$ with $C_2$" instead, or is my understanding even worse than I realize? – The Count Apr 05 '17 at 14:06
  • @TheCount Actually, it should read "external semidirect product." The direct product of $C_2$ and an abelian group is always abelian, and so would make the equation $ghg^{-1}=h^{-1}$ impossible! I'll amend the answer. – Stella Biderman Apr 05 '17 at 14:09
  • ^^ Well, it would be impossible for $ord(g)>2$ – Stella Biderman Apr 05 '17 at 14:16
  • OK, so would it be helpful for me to add a definition/explanation of what $\mathbb{Z}G$ is, or would that not be helpful? – The Count Apr 05 '17 at 14:19
  • @TheCount I'm trying to elucidate the situation for you. If you understand the connection to the group ring $\mathbb{Z}G$, then that's enough for me. – Stella Biderman Apr 05 '17 at 14:24
  • Ok, good. Your answer definitely made me feel better since you agree that the definition is awful. I guess I would feel a little better getting an answer one way of the other to "Is it necessary that $g\in G$? But I think I might be asking the wrong thing... thank you again! – The Count Apr 05 '17 at 14:29
  • @TheCount I've amended my answer to address that directly – Stella Biderman Apr 05 '17 at 14:31
  • Oh, wonderful! Thank you again so much. Boy, with your definition it is very clear, but theirs is just a catastrophe. – The Count Apr 05 '17 at 14:32
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    @StellaBiderman FYI, I showed this to my advisor today (we hadn't been able to touch base in about a week) and he said 1) That I am right (as are you) and $g\in G$, and 2) That indeed the definition is horribly presented and he likes the one you gave a lot more. So thanks again! – The Count Apr 05 '17 at 19:35