The exercise is as follows:
Let $A \lhd G$ and suppose $A = C_G(a)$ for every $a \neq 1$, $a \in A$. Assume further that $G/A$ is abelian. Show that $G$ has exactly $(|A| – 1)/[G:A]$ nonlinear irreducible characters, that all of these have degree equal to $[G:A]$ and vanish on $G \setminus A$. Assume $A < G$
He also gives a couple of hints. The first one is to bound $k = |\operatorname{Irr}(G)|$ from above by counting classes, in order to get
$k \leq 1 + \dfrac{|A|-1}{[G:A]} + \dfrac{|G|-|A|}{|A|}$
and the second one is to use characters to get
$k \geq [G:A] + \dfrac{|G|-[G:A]}{[G:A]^2}$
Since both of these quantities are actually the same, this would yield an equality.
My attempt
Notice, first, that, if $a \neq 1$ is in $A$, then $|a^G| = [G:A]$. Also, since $A \lhd G$, we obtain $a^G \subset A$. Therefore, if $n$ is the number of nontrivial conjugacy classes in $A$, then $1 + n[G:A] = |A| \iff n = \dfrac{|A|-1}{[G:A]}$. All we need to do now is determine the number of conjugacy classes outside of $A$.
For this, since $G/A$ is abelian, we get, given some $g \not \in A$, that $h^{-1}gh \in gA$, for all $h \in G$. In particular, $|g^G| \leq |gA| = |A|$ for all $g \not \in A$. Taking $m$ to be the number of such conjugacy classes, we get:
$$|G| – |A| = \sum_{i=1}^m |g_i^G| \leq m|A| \implies m \geq \dfrac{|G| – |A|}{|A|}$$
This would mean $k = 1 + n + m \geq 1 + \dfrac{|A|-1}{[G:A]} + \dfrac{|G|-|A|}{|A|}$, which is the opposite of what I'm supposed to be showing…
The other inequality was easy enough: there are at least $[G:A]$ linear characters lifted from $G/A$ and the equation $\sum \chi(1)^2 = |G|$ gives the result.
In short:
Where is my mistake in the proof of the first inequality?
It seems to me like there really is a possibility for there to be multiple "small" conjugacy classes outside of $A$…
Thanks in advance!
Best Answer
All references here are to Marty Isaacs' book Character Theory of Finite Groups.
What you showed @Gauss is not wrong, in fact (as @Brauer Suzuki pointed out), $|g^G|=|G:C_G(g)| \geq |AC_G(g):C_G(g)|=\frac{|A||C_G(g)|}{|C_G(g)|}=|A|$ for every $g \notin A$. Hence, together with your argument we see that in fact $m=\frac{|G|-|A|}{|A|}$. And $|g^G|=|A|$ for every $g \notin A$.
This implies that $A=G'$ ($G/A$ is abelian so $G' \subseteq A$ and the cardinality of any conjugacy class is always $\leq |G'|$).
(By the way, since $A \cap C_G(g)=1$ for every $g \notin A$, it follows that $Z(G)=\bigcap_\limits{g \in G} C_G(g)=1$. Also, $|A| \equiv 1$ mod $|G:A|$; of course, $A$ is abelian, so $G''=1$, that is, $G$ is metabelian.)
So, at this point we know that $G$ has $|G:G’|=|G:A|$ linear characters. And that the class number is $k(G)=1+ \frac{|A|-1}{|G:A|} + \frac{|G|-|A|}{|A|}=|G:A|+\frac{|A|-1}{|G:A|}$. We conclude that $G$ has $\frac{|A|-1}{|G:A|}$ non-linear characters. Owing to Problem $(2.9)$b $\chi(1) \leq |G:A|$ for every $\chi \in Irr(G)$. So with the formula $|G|=\sum_\limits{\chi \in Irr(G)}\chi(1)^2 \leq |G:A|+\frac{|A|-1}{|G:A|}|G:A|^2$, we see that in the latter equation one must have equality, implying that all the non-linear characters have degree $|G:A|$. Lemma (2.29) now shows that all these characters must vanish outside $A$, as wanted.