# 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 $|H_3|={p_1}^3$

• What have you tried so far ? I guess $C_G$ stands for the centralizer and $N_G$ the normalizer in $G$ – marwalix Jun 14 '15 at 6:45
• I think at least it's not hard to prove every Sylow $p$-group is abelian. – Censi LI Jun 14 '15 at 7:46
• i porved that. but i can't find relation between them an $G$ – erfan soheil Jun 14 '15 at 7:51
• Assuming that you have proved that the Sylow $p$-subgroups are all abelian, the result follows from Burnside's Transfer Theorem. – Derek Holt Jun 14 '15 at 10:33
• Note also that this is false in general for infinite groups. Counterexamples include nonabelian free groups and Tarski Monsters. – Derek Holt Jun 14 '15 at 13:31

$G$ 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.