Estimates on conjugacy classes of a finite group.

Question

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?

Answer 1

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 $|A|$ elements. 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 $\chi.$

Answer 2

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.