Normalizer and centralizer of abelian subgroups of a group are equal I have a question:

Let $G$ be a finite group. If for each abelian subgroup $H$ of $G$ the centralizer and the normalizer of $H$ are equal, that is, $C_G(H)=N_G(H)$, prove that $G$ is abelian group.

 A: First note that if $G$ has the property that the normalizer of each abelian subgroup equals its centralizer, then also all subgroups of $G$ have this property. So by induction we may assume that all proper subgroups of $G$ are abelian.
If $G$ is not a $p$-group then all its Sylow subgroups are abelian, and because of the above property they all have a normal complement by Burnside's transfer theorem. But then $G$ is nilpotent and hence - as a direct product of its abelian Sylow subgroups - abelian, too.
So we may assume that $G$ is a $p$-group. Take any maximal subgroup $H$ of $G$, and any element $g \in G \setminus H$. Because of maximality $H$ is normal in $G$, so $G$ centralizes $H$ and in particular $g$ commutes with every element of $H$. Since $H$ is abelian $G = \langle H, g \rangle$ is abelian, too.
A: I have learnt this proof from "A group-theoretic proof of a theorem of Maclagan-Wedderburn" by Zassenhaus. I typed it up in LaTeX a while ago, so here it goes:
Theorem. Let $G$ be a finite group. If $N_G(H)=C_G(H)$ for all abelian subgroups $H\subset G$, then $G$ is abelian.
Proof. We proceed by induction on the number of elements in $G$. Let $G$ be a finite group of order $n$ with the property that $N_G(H)=C_G(H)$ for all abelian subgroups $H\subset G$. We suppose that the theorem holds for all finite groups of order strictly less than $n$. Then, every proper subgroup of $G$ has order less than $n$, and satisfies the condition on the equality of normalizer and centralizer for abelian subroups, and so by induction hypothesis it is abelian. Thus, every proper subgroup of $G$ is abelian. Now, we divide the proof into two cases, whether or not the center of the group $Z=\{z\in G: ag=ga \ \forall g\in G\}$ is trivial.
Case 1: $|Z|>1$, i.e. the center of $G$ is not trivial. 
Then the quotient group $G/Z$ has order less than $n$. We will show that $G/Z$ is an abelian group. Assume $\overline{H}$ is any abelian subgroup of $G/Z$. We may assume $\overline{H}$ is a proper subgroup of $G/Z$, for otherwise $G/Z=\overline{H}$ is abelian, as desired. By correspondence theorem, there exists a subgroup $H\subset G$ containing $Z$ such that $H/Z=\overline{H}$. It is clear that $H$ is a proper subgroup of $G$ (because otherwise if $H=G$, then $\overline{H}=H/Z=G/Z$, but $\overline{H}$ is a proper subgroup of $G/Z$, contradiction). As we observed in above paragraph, properness of $H$ in $G$ together with induction hypothesis forces $H$ to be abelian. We will now prove that $N_{G/Z}(\overline{H})=C_{G/Z}(\overline{H})$. The inclusion $C_{G/Z}(\overline{H})\subset N_{G/Z}(\overline{H})$ is trivially true. For the reverse inclusion, let $\overline{X}$ be any element of $N_{G/Z}(\overline{H})$. Then, $(\overline{X})\overline{H}(\overline{X})^{-1}=\overline{H}$. If $\pi: G\to G/Z$ is the natural projection map, then for every $x\in\pi^{-1}(\overline{X})$, we have
$xHx^{-1}=H$. So $x$ lies in the normalizer of $H$, and since $H$ is abelian group, $x$ lies in the centralizer of $H$. But then, $xhx^{-1}=h$ for all $h\in H$. Applying the natural projection map $\pi$, we get $\overline{x}\overline{h}\overline{x}^{-1}=\overline{h}$. This is true for all $x\in\pi^{-1}(\overline{X})$. Consequently, $x$ lies in the centralizer $C_{G/Z}(\overline{H})$. We deduce that $N_{G/Z}(\overline{H})=C_{G/Z}(\overline{H})$. As $H$ was arbitrary abelian subgroup of $G/Z$, and $|G/Z|<n$ by induction hypothesis it follows that $G/Z$ is abelian, as desired. We will now show that $G$ must also be abelian. Let $a,b\in G$ be any two arbitrary elements. By definition of center, $bz=zb$ for each $z\in Z$. Furthermore, $Z$ itself is clearly an abelian group; hence the subgroup generated by $(b,Z)=(b,Z)$ is an abelian subgroup of $G$. Since $G/Z$ is abelian, we have 
$$
(aba^{-1}Z)=(aZ)(bZ)(a^{-1}Z)=(bZ)(aZ)(a^{-1}Z)=bZ(aa^{-1}Z)=bZ
$$
So $aba^{-1}Z=bZ$. Thus, there exists $z\in Z$ with $aba^{-1}=bz$. Since $za=az$ for all $z\in Z$, it is clear that if $y\in \langle b,Z\rangle$, then $aya^{-1}\in \langle b,Z\rangle$. Consequently, $a\langle b,Z\rangle a^{-1}=\langle b,Z\rangle$, meaning that $a$ is the normalizer of $\langle b,Z\rangle$, and so $a$ is in the centralizer of $\langle b,Z\rangle $ (because $\langle b,Z\rangle $ is an abelian subgroup of $G$). In particular, $a$ commutes with $b\in \langle b,Z\rangle$, giving us $ab=ba$. As $a,b\in G$ were arbitrary, we deduce that $G$ is abelian.
Case 2: $|Z|=1$, i.e. the center of $G$ is trivial. 
Assume $M_1$ and $M_2$ are two distinct maximal proper subgroups of $G$. By induction hypothesis, $M_1$ and $M_2$ are abelian. Also, the subgroup generated by $M_1$ and $M_2$ must be the entire group, i.e. $\langle M_1,M_2\rangle=G$ (otherwise, maximality of $M_1$ would be contradicted). Now, if $x\in M_1\cap M_2$, then $x$ commutes with every element of $M_1$ and $x$ commutes with every element of $M_2$. Since $M_1$ and $M_2$ are abelian, it follows that $x$ commutes with every element of $\langle M_1, M_2\rangle=G$. As a result, $x\in Z=\{1\}$. We conclude that any two distinct maximal subgroups of $G$ intersect at the identity element. Now suppose $M$ is any maximal proper subgroup of $G$. We will now analyze conjugate subgroups of $M$. First, for any $x\in G$, the conjugate subgroup $xMx^{-1}$ is maximal (Indeed, if $xMx^{-1}\subsetneq Q$ then $M\subsetneq x^{-1}Qx$, and so $x^{-1}Qx=G$, implying that $Q=G$). Suppose $xMx^{-1}=yMy^{-1}$ for some $x,y\in G$. Then, $(y^{-1}x)M(y^{-1}x)^{-1}=M$, and so $y^{-1}x$ is in the normalizer of $M$, and hence, in the centralizer of $M$ (because $M$ is an abelian subgroup of $G$). So $y^{-1}x$ commutes with every element of $M$, and consequently commutes with every element of $\langle y^{-1}x, M\rangle$. If $y^{-1}x\notin M$, then by maximality of $M$, $\langle y^{-1}x, M\rangle = G$, and so $y^{-1}x$ commutes with every element of $G$, i.e. $y^{-1}x\in Z$, so $y^{-1}x$ is the identity element (as we are in Case 2), and  so $y^{-1}x\in M$ contradiction, as we were assuming $y^{-1}x\notin M$. Thus, whenever $xMx^{-1}=yMy^{-1}$, it follows that $y^{-1}x\in M$. The converse is also true: if $y^{-1}x\in M$, then $(y^{-1}x)M(x^{-1}y)=M$, and so $yMy^{-1}=y(y^{-1}xMx^{-1}y)y^{-1}=xMx^{-1}$. So, $xMx^{-1}=yMy^{-1}$ if and only if $y^{-1}x\in M$. Equivalently, $xMx^{-1}=yMy^{-1}$ if and only if $xM=yM$. We deduce that the number of conjugates of $M$ is equal to the number of left cosets of $M$, i.e. there are $|G|/|M|$ conjugates of $M$. Since any two distint maximal subgroups of $G$ intersect at identity (as we proved above), and since each conjugate subgroup contains $|M|-1$ non-identity elements, the total number of non-identity elements in all conjugates of $M$ is 
\begin{equation}
\frac{|G|}{|M|}(|M|-1)=|G|-\frac{|G|}{|M|}  
\end{equation}
We now have enough tools to show $G$ is abelian. Take a non-identity element $x\in G$. If $(x)=G$, then $G$ is cyclic, and hence abelian. We may assume $(x)\subsetneq G$, so that $(x)$ is contained in some maximal subgroup $M\subset G$. Since $|M|>1$ (because $(x)\subset M$ and $|(x)|>1$), we see from equation $(1)$ that $M$ and its conjugates supply $|G|/2$ non-identity elements of $G$. If $G$ contained any element $w$ that is not in $M$ or in any of the conjugates, then $w$ would be contained in some other maximal subgroup, say $P\subset G$. Since any two distinct maximal subgroups of $G$ are distinct, $P$ and its conjugates supply $|G|/2$ non-identity elements of $G$, all of which all of which are different from the above $|G|/2$ non-identity elements supplied by $M$ and its conjugates. Thus, $P$, $M$ together with conjugates supply $|G|/2 + |G|/2= |G|$ non-identity elements of $G$. Adding the identity element, we get a total of $|G|+1$ elements in $G$, contradiction. Hence, $G$ cannot contain any element $w$ that is not in $M$ or any of its conjugates. We conclude that $M$ and its conjugates supply all the elements of $G$. From equation $(1)$, this means
$$
|G|-\frac{|G|}{|M|} + 1 = |G|
$$
giving us $\frac{|G|}{|M|}=1$, so that $G=M$. This contradicts the fact that $M$ is a proper subgroup of $G$, and establishes the theorem.
