Show that the commutator subgroup of an abelian normal subgroup and whole group has special form. Let $A$ be an abelian normal subgroup and $x \in G$. Let $G = AC_G(ax)$ for all $a \in A$, then
$$
 [A, G] = \{ [a,x] : a \in A \}.
$$
Any hints?
EDIT: What I have done so far. Let $y = [a, g] \in [A, G]$ be a generator of $[A,G]$, then I can choose some $a'$ and by the assumptions there is some decomposition $g = \hat{a} z$ with $z \in C_G(a'x)$, i.e. $1 = [z, a'x] = [a'x, z]$. Further
$$ 
 y = [a,g] = a^{-1} g^{-1} a g = a^{-1} z^{-1} \hat{a}^{-1} a \hat{a} z 
   = a^{-1} z^{-1} a z = [a,z]
$$
by knowing that $A$ is abelian. I have to show that $[a,z] = [a'', x]$ for some $a'' \in A$. I guess I have to choose $a'$ in some clever way, but I do not see anything. Some other things I know:
i) $[a,g] \in A$ for each $a \in A, g \in G$ by normality,
ii) the map $a \mapsto [a,x]$ from $A \to A$ is a homomorphism,
iii) $[A, \langle x \rangle] = \{ [a,x] : a \in A \}$.
 A: Let $H=\{[a,x]:a\in A\}$.  First, let's show that the normal subgroup generated by $H$ contains $[A,G]$.  That is, we want to show that if we apply any homomorphism $\varphi$ that kills $H$, $\varphi(a)$ commutes with $\varphi(y)$ for any $a\in A$, $y\in G$.  To show this, write $y=bz$, where $b\in A$ and $z$ commutes with $ax$.  Since $a$ commutes with $b$, it suffices to show $\varphi(a)$ commutes with $\varphi(z)$.  Now write $z=cw$, where $c\in A$ and $w$ commutes with $x$.  Since $\varphi$ kills $H$, $\varphi(c)$ commutes with $\varphi(x)$, so this means $\varphi(z)$ commutes with $\varphi(x)$.  Since $z$ commutes with $ax$, this now implies $\varphi(z)$ commutes with $\varphi(a)$, as desired.
It now suffices to show that $H$ is a normal subgroup.  The fact that $H$ is a subgroup follows from your observation (ii).  For normality, fix $a\in A$ and $y\in G$ and write $y=bz$ where $b\in A$ and $z$ commutes with $x$.  Then $$y[a,x]y^{-1}=b[zaz^{-1},zxz^{-1}]b^{-1}=b[zaz^{-1},x]b^{-1}.$$
Since $A$ is normal $zaz^{-1}\in A$, and hence $[zaz^{-1},x]\in A$ as well, so it commutes with $b$.  Thus $y[a,x]y^{-1}=[zaz^{-1},x]\in H$.  Since $y\in G$ and $a\in A$ were arbitrary, $H$ is normal.
A: Here is another solution. By putting  $a'=1$ in your calculation, you get $[a,g]=[a,g_1]$ with $g_1 \in C_G(x)$.
Since $G = AC_G(ax) = C_G(ax)A$, there exists $a' \in A$ with $g_1a' \in C_G(ax)$. So, using $[a,a']=[x,g_1]=1$, we get, using the commutator laws:
$$1 = [ax,g_1a'] = [a,g_1a']^x[x,g_1a'] = [a,g_1]^x[x,a'].$$
Conjugating by $x^{-1}$ then gives $1=[a,g_1][x,a'']$, with $a'' = (a')^{x^{-1}}$, so $[a,g] = [a,g_1] = [a'',x]$, and we are done.
