Suppose that $G$ is a finite group and not simple,and $H$ is a minimal normal subgroup of $G$,and also $P$ is a Sylow $p$-subgroup of $H$.
if $P$ is abelian and $\frac{N_{G}(P)}{P}$ will be a group of order prime number $q$ then $P=H$ .
my solution:I think it is enough to show that $P$ is normal complement in $H$,then because $N$ is minimal normal subgroup of $G$ then it is trivial minimal normal subgroup . and then $P=H$
but I don't know how to show that $P$ is normal complement in $H$.
it will be great if you help me with this,thanks.
Since $P$ is abelian, $P \subseteq C_G(P) \subseteq N_G(P)$. So if index$[N_G(P):P]$ is prime, then one case is $N_G(P)=C_G(P)$, so certainly $N_H(P)=C_H(P)$. You can now apply Burnside's Theorem in the subgroup $H$ (see for instance M.I. Isaacs, Finite Group Theory, 5.13): $P$ has a normal complement in $H$. Since $H$ is a minimal normal subgroup, it then follows that $P=H$.
So we are left with analyzing the case $P=C_G(P)$ and index$[N_G(P):P]$ is prime.
We are going to use the Frattini Argument: in any case $G=HN_G(P)$. Note that $P \subseteq N_G(P) \cap H \subseteq N_G(P)$. So either $P=N_G(P) \cap H$, or $N_G(P) \cap H=N_G(P)$. In the latter case $N_G(P) \subseteq H$, and it follows that $G=H$, a contradiction ($G$ is not simple).
So assume $P=N_G(P) \cap H$, hence $P=N_H(P)$, and we also had $P=H \cap C_G(P)=C_H(P)$. Again we can apply Burnside's Theorem to arrive at $P=H$.