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Following on from some of myprevious MO questions on finite group theory... $\newcommand{\Irr}{\operatorname{Irr}}\newcommand{\Conj}{\operatorname{Conj}}\newcommand{\AMZL}{{\rm AM}_{\rm Z}}$

Let $G$ be a finite group, $\Irr$ the set of irreducible characters (working over $\mathbb C$) and $\Conj$ the set of conjugacy classes. Consider the following quantity, which seems to have first been introduced in work of Azimifard, Samei and Spronk $$\alpha_G = \frac{1}{|G|^2} \sum_{\phi\in\Irr} \sum_{C\in\Conj} \phi(e)^2 |\phi(C)|^2 |C|^2 . $$ It is not hard to show, just using basic character theory, that $\alpha_G \geq 1$ with equality if and only if $G$ is abelian. This constant/invariant arises as a minorant for a more interesting invariant of $G$ which was studied in the aforementioned paper.

Call this "more interesting invariant" $\AMZL(G)$. It turns out that $\AMZL(G)\geq \AMZL(G/N)$ for every normal subgroup $N$, a property which is vital to some work I am writing up that obtains lower bounds on $\AMZL(G)$. One branch of the case-by-case attack I'm using works by getting lower bounds on $\alpha_G$ when $G$ has trivial centre, and so if $\alpha_G$ never increases when we replace $G$ with a quotient then I could in fact get general lower bounds on $\alpha_G$ by inductively modding out by the centre.

Question. Do there exists a finite group G and a normal subgroup N for which $\alpha_{(G/N)} > \alpha_G$ ?

I did try for a while to show this can never happen but rapidly got stuck, so I'm hoping that MO readers who are more skilled/experienced with finite groups can either suggest counterexamples to try, or give some evidence that no such counterexamples exist. Please bear in mind that I have next to no experience with GAP or similar, so I don't mind concrete suggestions of code, but responses of the form "get GAP to look through all non-simple non-abelian examples of order < 100" may not be as helpful as their authors imagine.


Just in case it helps to rule out certain avenues of attack: if $G$ is a group with two character degrees (i.e. there is an integer $m$ such that $\phi(e)\in \{1,m\}$ for all $\phi\in\Irr$) then one can evaluate $\alpha_G$ explicitly, using ideas similar to this paper: one obtains

$$ \alpha_G - 1 = (m^2-1) \left( 1- \frac{|L|}{|G|^2} \sum_{C\in\Conj} |C|^2 \right) $$ where $|L|$ is the number of linear characters on $G$.

In particular, suppose if $G=H \rtimes C_2$ is a generalized dihedral group with $|H|=2n+1$ ($n$ a positive integer). Back of the envelope scribbling gives me $m=2$, $|L|=2$ and the conjugacy classes have sizes $1$, $2$ repeated $n$ times, and $2n+1$, so that $$ \frac{\alpha_G -1}{3} = 1 - \frac{1}{2(2n+1)^2}(1+ 4n + (2n+1)^2) = \frac{1}{2}\left(1-\frac{1}{2n+1}\right)^2 $$ It also seems to me that proper quotients of $G$ must either be abelian or generalized dihedral with smaller order, in which case we would have $\alpha_G \geq \alpha_{(G/N)}$ for every $N \lhd G$.

share|improve this question
    
Something must be wrong here, because if $\alpha$ has the properties you describe then examples of $\alpha_G > \alpha_{G/N}$ must abound: given that $\alpha_G \geq 1$ with equality iff $G$ is abelian, it's enough to use any $G,N$ where $G/N$ is abelian but $G$ is not. For example, $G=S_n$ and $N=A_n$ for any $n\geq 3$, or $G$ the quaternion group and $N$ its center. –  Noam D. Elkies Aug 14 '13 at 23:22
    
@NoamD.Elkies Did I mistype something? I'm looking for examples where alpha of the quotient is strictly greater than alpha of the original G. –  Yemon Choi Aug 14 '13 at 23:54
    
Sorry, my fault: I misread the inequality; you're right. –  Noam D. Elkies Aug 15 '13 at 1:34

1 Answer 1

up vote 4 down vote accepted

Let $G= \operatorname{SL}(2,5)$ and $N= \mathbf{Z}(G)$, so that $G/N = \operatorname{PSL}(2,5)\cong A_5$. If my code works correctly, then $$ \alpha_{G/N} = 842/75 > 6661/600 = \alpha_G. $$

My implementation of alpha (for character tables) in GAP:

alpha:= function( tbl )
    local clssizes, n, kg, phi, i;
    clssizes:= SizesConjugacyClasses( tbl );
    n:= Size( tbl );      # group order
    kg:= NrConjugacyClasses( tbl );
    return Sum( Irr( tbl ), 
                phi-> phi[1]^2 
                      * Sum( [1..kg], 
                             i-> phi[i] * ComplexConjugate( phi[i] )
                                        * clssizes[i]^2 
                           )
              )/ n^2 ;
end;
share|improve this answer
    
Thanks! I'll take your word for it that code works... –  Yemon Choi Aug 26 at 0:45
    
@YemonChoi: Well, at least it gives back $1$ for some small abelian groups I tested, and also the same results as in your question for small dihedral groups. –  Frieder Ladisch 2 days ago
    
That's good to know. I think that for the affine group of a finite field of order q one should get 3-4/q –  Yemon Choi 2 days ago

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