Let $A\lhd G$ for a finite, nontrivial nonabelian group $G$. Suppose that $A$ and $G/A$ are abelian. If for each $a\in A-\{1\}, C_G(a)=A$, then $A=G'$ I'm working on proving the following problem:
Let $A \lhd G$ for a finite, nontrivial, nonabelian group $G$. Suppose that $A$ and $G/A$ are abelian. If for each $a \in A - \{1\}$, $C_G(a) = A$, then $A = G'$.
I can't seem to prove this.  The way I've been trying to prove this is to assume that $G' \subsetneq A$, and using the fact that $C_G(a) = A$ to force a contradiction by showing that $G/G'$ is nonabelian, but nothing I'm trying is working. Maybe there is a direct way to show that each element of $A$ is a commutator of some elements of $G$. Any suggestions would be appreciated.
 A: This is a detailed solution.
We use the notations $a^b=b^{-1}ab$ and $[a,b]=a^{-1}a^b$.

*

*If $a\in A$, $b\notin A$, and $m$ is the order of element $b$ then $$
a'=a\cdot a^b\cdots a^{b^{m-1}}=1.
$$
Indeed, since $A$ is abelian, $a'^b=a'$ and hence $b\in C_G(a')$. If $a'\neq1$, then $b\in C_G(a')=A$.


*If $a\cdot a^b\cdots a^{b^{m-1}}=1$, then $a^m\in G'$.
In fact we have
$$
1=a\cdot a^b\cdots a^{b^{m-1}}=a\cdot(a[a,b])\ldots(a[a,b^{m-1}])=a^m[a,b]\ldots[a,b^{m-1}].
$$


*If $b\notin A$ and $b^s\in A$, then $b^s=1$ (otherwise since $b\in C_G(b^s)$ and this is by convention impossible).


*For any prime $p$ if $p$ is divisor $|A|$, then any Sylow $p$-subgroup of group $G$ lies in $A$.
Let us prove it. Let $P$ be a Sylow $p$-subgroup.
Since $A$ is a normal subgroup in $G$, $p$ is a divisor of $|A|$, and $P$ is Sylow $p$-subgroup it follows that $B=P\cap A\neq\{1\}$, and then $Z(P)\cap B\neq\{1\}$. Let $a\in Z(P)\cap B$, $a\neq1$. We obtain that $P\leq C_G(a)=A$.
Our statement now follows from 1-4.
If $p$ is prime and $p$ is divisor $|G/A|$, then there exists $bA\in G/A$ of order $p$ $\Rightarrow b^p\in A\Rightarrow b^p=1$ (see 3).
By virtue of 4, $p$ does not divide $|A|$.
It follows from 1 and 2 that $a^p\in G'$ for every $a\in A$. Since $(p,|A|)=1$, then $\{a^p\mid a\in A\}=A$, so $A\leq G'$, that is $A=G'$.
A: I have came up with an answer to my own question involving characters.
We are going to bound the number of irreducible characters of $G$ in two different ways. First by counting the number of conjugacy classes. Since each nonidentity element of $A$ has centralizer $A$, the size of the conjugacy class to which any $a \in A - \{1\}$ belongs is $|G:A|$. So the number of conjugacy classes of $G$ contained in $A$ is $1 + (|A| - 1)/|G:A|$. Now, since no element of $G - A$ centralizes a nonidentity element of $A$, it follows that for a fixed  $g \in G - A$, each conjugate of $g$ by an element of $A$ is distinct, so each conjugacy class contained in $G - A$ has at least $|A|$ elements. Since the number of irreducible character of $G$ is the number of conjugacy classes of $G$, it follows that the number of irreducible characters is less than $$1 + (|A| - 1)/|G:A| + (|G - A|)/|A| $$ $$= (|A| - 1)/|G:A| + |G:A|$$
Now, we bound the number of irreducible characters from below using characters. We use the following facts:

*

*$|G| = \sum_{\chi} \chi(1)^2$ where the sum ranges over all irreducible characters of $G$.

*The number of linear characters of $G$ is precisely $|G:G'|$.

*Since $A$ is an abelian subgroup, $\chi(1) \leq |G:A|$ for each irreducible character $\chi$ of $G$.

Since $G/A$ is abelian, $G' \leq A$, and the number of linear characters of $G$ is greater than $|G:A|$. Let $n$ denote the number of irreducible characters of $G$.
$$
n = |G:A| + \sum_{\chi} 1
$$
where $\chi$ in the sum ranges over all irreducible characters of $G$ except exactly $|G:A|$ linear characters. Now, $\chi(1) \leq |G:A|$ for each irreducible character $\chi$, so we have that
$$
n \geq |G:A| + \dfrac{1}{|G:A|^2}\sum_{\chi} \chi(1)^2 = (|G| - |G:A|)/|G:A|^2 $$
$$ = (|A| - 1)/|G:A| + |G:A| $$
This lower bound is the same as the upper bound found for the number of irreducible characters of $G$ so the inequality above is actually equality. This equality can only happen when $\chi(1) = |G:A|$. If $A$ is a proper subgroup, then each $\chi$ in the sum has $\chi(1) > 1$, and the only linear characters are the $|G:A|$ ones excluded from the sum.
We have shown that the number of linear characters of $G$ is $|G:A|$, so $|G'| = |A|$, but $G' \subseteq A$ since $G/A$ is abelian, so $G' = A$.
