If $C_G(H)=N_G(H)$ for all abelian subgroups, prove that $G$ is abelian Let $G$ be a finite group such that for all abelian subgroups $H$ of $G$,
$$C_G(H)=N_G(H).$$
Prove that $G$ is abelian.
($C_G(H)$ is the centralizer, $N_G(H)$ is the normalizer of $H$ in $G$)
my try : let $G={p_1}^{\alpha_1}\dots p_n^{\alpha_n}$, then $G$ has an subgroup of prder $p_1$ like $<a> (a \in G)$ such that $<a> \unlhd H_2$ , where $|H|=p_1^2$. since $<a>$ is abelian so $$C_G(<a>)=N_G(<a> )=H_2 $$
then $H_2$ is an abelian subgroup suchthat $H_2 \unlhd h_3 $ where $|H_3|={p_1}^3$ 
 A: $G$ has the property that for each abelian subgroup $A$ of $G$, $C_G(A)=N_G(A)$. Note that this property carries over to subgroups: if $H \leq G$, and $B$ is an abelian subgroup of $H$, then $B$ is certainly an abelian subgroup of $G$ and $C_H(B)=H \cap C_G(B)=H \cap N_G(B)=N_H(B)$. So, by induction on the order of $G$, if $G$ is not a $p$-group for some prime $p$, all Sylow subgroups are proper and hence abelian, with their centralizer and normalizer being equal. Now we follow the hint of Derek Holt: apply Burnside's Theorem (the proof requires some sophisticated group theory - transfer, see for example I.M. Isaacs, Finite Group Theory, Theorem 5.13): if $P \in Syl_p(G)$, then $G=PN$, with $N \lhd G$ and $N \cap P=1$. Here $P$ is abelian and since $N$ is proper (whence abelian), induction gives $C_G(N)=N_G(N)=G$. Hence $N \subseteq Z(G)$ and $G=NP$ must be abelian.
Now assume that $G$ is a $p$-group for some prime $p$. Let $A$ be maximal among the abelian normal subgroups of $G$. Such an $A$ exists since $Z(G)$ is non-trivial. We may assume that $A \lneq G$, otherwise we are done. We argue that $A=C_G(A)$ to arrive at a contradiction: since $G$ is a $p$-group, normalizers grow, that is, $A \lneq N_G(A)=C_G(A)$ by hypothesis.
So for the final argument write $C=C_G(A)$ and suppose $A \lneq C$. It is easy to show $C$ is normal in $G$. Now by standard $p$-group theory (see Lemma 1.23 in the aforementioned book), we can find $B \unlhd G$, with $A \leq B \leq C$, and index$[B:A]=p$. Since $B \subseteq C$, we have $A \subseteq Z(B)$. Hence $B$ must be abelian (apply that if $|B/Z(B)| \text{ divides }p$, then $B$ is abelian). This contradicts the maximality of $A$ and the proof is finished.
