Let $G$ be a nonabelian $p$-group and $H$ a minimal nonabelian $p$-subgroup of $G$. In some books (see for exemple Remark 1, p. 331 in the Book" Berkovich, Groups of prime power ordre, Vol. 3" ), it is deduced directly that $C_{G}(H)=Z(H)$ without any clarification. But I can't understand why this is true. Did I miss something?
I would appreciate any hints and comments. Thank you in advance!
Berkovich-Janko do say this, but they have made a leap. In this situation, $G$ is a non-abelian $p$-group of order $>p^4$, and all proper non-abelian subgroups of $G$ have centres of order $p$. Thus if $B$ is minimal non-abelian then $Z(B)$ has order $p$. Now, if $x\in C_G(B)$ then $H=\langle x\rangle\cdot B$ is a non-abelian subgroup and $x\in Z(H)$. But $Z(B)\leq Z(H)$, so $x\in Z(B)$. Thus $C_G(B)\leq B$.