Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Initial caveat: the following question could probably be answered by Google, MathSciNet or my library, if I could find the right search terms or book... but I've not had any luck today, so I hope someone can point me to a reference.

(The question is related to some of my older questions concerning characters of finite groups. All representations/characters are over the complex field.)

I am trying to estimate a certain invariant associated to finite groups, and recently thought that a useful toy example to play with would be $$ G = \left\{ \left( \begin{matrix} a & b \\ 0 & 1 \end{matrix} \right) \;\mid\; a,b\in {\mathbb Z}/p^n{\mathbb Z}, p \nmid a \right\} $$ where $p$ is a prime and $n\geq 1$. I guess this might be called the "affine group" of the ring $R={\mathbb Z}/p^n{\mathbb Z}$?

Now the invariant can be calculated pretty easily (modulo tedious sums) once we know the character table of $G$, but this means more than knowing the degrees of the irreducible characters; I need to know the values they take on the various conjugacy classes inside $G$.

For $n=1$ this does not take long to do directly and can also be found as an example in various introductory-level textbooks on representation theory. However, for $n=2$ the best I could find was a section in a paper of several authors, where they just work out the character table by hand after first finding the characters via induction ($G$ is a semi-direct product arising from the action of the group of units in $R$ on the additive group of $R$). Now since I want to continue to higher $n$, I seem to be faced with three options:

1) Slog through the computation myself (which is probably good for my mathematical soul, but takes time & brainpower I need to spend on other things)

2) Learn how to ask a computer to do this (see previous parenthetical remark)

3) Find a reference which just gives the table.

So before embarking on 1), I thought I'd ask here. Most sources I could find from a crude skim online and in my library only discussed linear groups over finite fields; but I'm hoping that the construction here is sufficiently natural that it might have been treated already and written up somewhere.

share|improve this question
2  
Do what's good for your soul, is my advice. Work out the representations, not just the characters. –  Tom Goodwillie Oct 30 '10 at 2:11
    
I see Denis has changed my \nmid to \not\mid -- which one works better, in people's experience? (When I was typing this question, \not\mid gave a rather ugly result.) –  Yemon Choi Oct 30 '10 at 7:58
    
@Yemon: The \not\mid looked terrible to me. So, being bold (wait, that's not our motto! but anyway) I changed it back to \nmid to see what that looks like. My answer: it looks much better. –  Pete L. Clark Oct 30 '10 at 8:03
    
@Pete: I am still in the process of working out the inner workings of this site, but could you not just have rolled back Denis's change? I understand that now, a revision by one more person will turn the post into CW, which would do Yemon injustice. Or is rolling back also counted as a third edit? –  Alex B. Oct 30 '10 at 8:12
    
@Alex: I didn't roll back because I (wrongly) thought that Denis had made some other change in his edit [the comment threw me off a bit]. So I agree that I could have and probably should have. However, if the post becomes CW, which hasn't happened yet, I would hesitate to call it "injustice". After all, this is a Q&A site, he asked a question, and he has already received an excellent answer. That's justice, right? –  Pete L. Clark Oct 30 '10 at 8:24
show 4 more comments

1 Answer

up vote 5 down vote accepted

The groups you are interested in are sometimes called false Tate extensions in number theorists' jargon. They are Galois groups of the Galois closures of extensions of $\mathbb{Q}$ obtained by adjoining the $p^n$-th roots of a $p$-th power free element. The irreducible representations are very explicitly described in Vladimir Dokchitser's PhD thesis, beginning of chapter 3.

Alternatively, the characters, together with their values, are easy to compute using a procedure described in Serre's Linear representation book, part II, Section 8.2. He explains how to obtain all the irreducible characters of any group that is a semi-direct product when the normal subgroup is abelian. Note that the character of an induced representation is easy to compute in terms of the original character and the coset-action.

And of course one more remark: for any given $p$ and $n$, MAGMA will just give you the character table.

Edit: Since you seem unsure, how to ask a computer for the character table, here is MAGMA code as an example. You will easily adopt it to any other package that handles character tables:

p:=3; n:=2;
Z:=Integers();
gl:=GL(2,quo<Z|p^n*Z>);
pr:=PrimitiveRoot(p^n);

G:=sub<gl|gl![[1,1],[0,1]],gl![[pr,0],[0,1]]>;
CharacterTable(G);
share|improve this answer
    
Thanks, Alex. I tried to make it clear in the question that I am aware you can work out the characters by inducing up from the abelian normal subgroup (you have a small Fingerfehler there) but that, since I don't do this kind of thing as part of my day-to-day teaching or research, I'd be slow and unreliable at it. Your link was helpful and I will also check out Serre's book; but given that knowing the full character table is only the start of the computation I wish to do, perhaps MAGMA is the way to go for me. –  Yemon Choi Oct 30 '10 at 1:55
    
So you did. Sorry, I only skim-read your post before I enthusiastically started typing :) At any rate, Vladimir Dokchitser should do 3 for you and the magma code will do 2. –  Alex B. Oct 30 '10 at 2:00
    
(Somewhat amusingly, my library appears not to have the GTM copy of Serre's book, but it does have the French original. Alors je n'ai pas reussi d'echapper la langue francaise...) –  Yemon Choi Nov 1 '10 at 4:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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