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In Character Theory Of Finite Groups by I Martin Issacs as exercise 2.18, on page 32.

Theorem:
Let $A$ be a normal subgroup of $G$ such that $A$ is the centralizer of every non-trivial element in $A$. If further $G/A$ is abelian, then $G$ has |G:A| linear characters, and $(|A|-1)/|G:A|$ non-linear irreducible characters of degree =|G:A| which vanish off $A$.
My Attempt:
By the hypotheses, every conjugacy class contained in $A$ has order=|G:A|, except the trivial one. Moreover, we find that if $C$ is a class which contains one element in $\alpha A$, then $C$ is contained in $\alpha A$. Let $A$ act on $C$ by conjugation and partition $C$ into orbits. Again we find that no element in $C$ is fixed by $A$, so that |C| is greater than |A|, thus
k=the number of classes in $G$ is $\le 1+(|A|-1)/|G:A|+(|G|-|A|)/|A|$.
On the other hand, as $G' \subset A$, we find that the number of linear characters is $\ge |G:A|$. Furthermore, by Mackey's irreducibility criterion, there are exactly (|A|-1)/|G:A| irreducible characters induced by linear ones of $A$. Therefore, we conclude as stated.

As is obvious, this approach, if correct, exploits properties of induced characters of Mackey, with which I am still not so familiar, and hence I might ask:
I: Is my try valid?
II:How to proceed in an elementary manner?

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Sorry if it is inappropriate to ask again and again questions about some exercises in one particular book. I tried really to resolve the problem, but obtained nothing other than the above attempt. Maybe I shall consult my teacher, but he is not interested enough in such things, and his course on representations appear not to contain some necessary properties here, so I cannot discuss with other attendants, who are as well busy. Thanks for any help then. –  awllower Jun 3 '12 at 18:01

2 Answers 2

up vote 5 down vote accepted

I do not think you need to use Mackey or other induction theorems. You have already observed that (counting the identity), there are exactly $1 + \frac{(|A|-1)}{[G:A]}$ conjugacy classes meeting $A.$ Now choose an element $b \in G \backslash A.$ Notice that $|C_{G}(b)| \geq [G:A]$ because there are are at least $[G:A]$ linear characters of $G$, as you have already noted, and for any linear character $\mu$ of $G$ we have $|\mu(b)|^{2} = 1.$ On the other hand, $C_{G}(b) \cap A = 1,$ so $|G| \geq |A| |C_{G}(b)|$ and $|C_{G}(b)| \leq [G:A].$ Hence $|C_{G}(b)| = [G:A]$ for each $b \in G \backslash A.$ Hence there are $[G:A]-1$ conjugacy classes of $G$ which do not meet $A$, and each of these contains $|A|$ elemments. Thus $G$ has $[G:A] + \frac{|A|-1}{[G:A]}$ conjugacy classes, hence the same number of complex irreducible characters. Since $|C_{G}(b)| = [G:A]$ for each element $b$ of $G \backslash A,$ there can be no more that $[G:A]$ linear characters of $G,$ so there are exactly $[G:A]$ such linear characters, as we know there are at least that many. Furthermore, from the orthogonality relations, we see that whenever $\chi$ is a non-linear irreducible character of $G$, we must have $\chi(b) = 0$ for all such $b.$ Also, we have $\chi(1) \leq [G:A]$ by other results in Isaacs book, so you have enough information to deduce that $\chi(1) = [G:A]$ for all such non-linear irreducible $\chi.$

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Really thanks for the help. This solution looks simple and powerful, while the most part is still beyond my reasoning. Per chance you could spend some time and expand the arguments further? I could not see why $|C_{G}(b)| \geq [G:A]$, which might therefore be a good start. In fact, if I could show the exact number of conjugacy classes, I could have finished the proof. So, thanks again. –  awllower Jun 4 '12 at 3:32
1  
I won't re-edit the answer, everything is there, but I'll say a little more here. There are at least $[G:A]$ linear characters, so these already contribute at least $[G:A]$ to $\sum_{\chi \in {\rm Irr}(G)} |\chi(b)|^{2}.$ Since the latter sum is $|C_{G}(b)|,$ we must have $|C_{G}(b)| \geq [G:A].$ –  Geoff Robinson Jun 4 '12 at 13:40
    
Oh!!I found out the reasoning behind this argument, or in particular the part concerning those centralizers and characters: almost all of it arise from that "second orthogonality relation", as in your comment!! This is really marvelous, thanks very much. –  awllower Jun 6 '12 at 4:52

Let $k$ be the number of conjugacy classes, which is the number of irreducible characters. Then as you said, we have $$ k\le 1+\dfrac{|A|-1}{[G:A]} + \dfrac{|G|-|A|}{|A|}.$$

Now there are at least $[G:A]$ linear characters. Now we know by problem 2.9 that for the nonlinear characters $\chi$, $\chi(1)\le [G:A]$. Let $\chi_1,\ldots,\chi_m$ be linear characters of $G$ (with $m=[G:A]$). Let $\chi_{m+1},\ldots,\chi_k$ be the other irreducible characters (perhaps some linear). We know $$ |G| = \sum_{i=1}^k\chi_i(1)^2,$$ or put another way, $$ |G| - m = \sum_{i=m+1}^k\chi_i(1)^2.$$

Since $\chi_i(1)\le m$ for all $i$, this yields $$ |G|-m\le (k-m)m^2$$ or $$ \frac{|G|-m}{m^2}+m\le k.$$

Replacing $m$ by $[G:A]$ gives $$ k \ge [G:A]+\frac{|G|-[G:A]}{[G:A]^2}.$$

Cleaning things up and using common denominators, we get $$ \dfrac{|G|[G:A]}{|G|}+\dfrac{|A|^2-|A|}{|G|}\le \dfrac{|G|}{|G|}+\dfrac{|A|^2-|A|}{|G|}+\dfrac{|G|[G:A]-|G|}{|G|}.$$

This is an equality, so we must have had equality throughout. In particular, there are exactly $[G:A]$ linear characters, and for nonlinear characters, $\chi(1)=[G:A]$. The vanishing outside $A$ follows from Lemma 2.29.

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I have tried but to no avail to deduce the second inequality in the answer, but I failed to find out the reason behind this, though simple inequality: $k \ge [G:A]+\dfrac{|G|-[G:A]}{[G:A]^2}$ It seems not true in general, which motivated my attempt to deploy the induction theorems; in any case, thanks for the help. –  awllower Jun 4 '12 at 3:28
    
I have added some details. –  user641 Jun 4 '12 at 4:39
    
Thanks, now I understand the whole matter. –  awllower Jun 5 '12 at 6:56

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