If $G'=G$, and suppose $N_G(P)$ is solvable for $P\subseteq G$, show that $ G$ is simple.

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Given a finite nontrivial group $G$, Suppose $N_G(P)$ is solvable for every nontrivial $P\subseteq G$ of prime power order. If $G'=G$, (where $G'$ is the derived subgroup) show that $G$ is simple.

I understand the pieces of information given in the problem but how to put them together to make an argument is challenging to me. I will appreciate any help.

I am thinking that, if I start by assuming $G$ is not simple and that there exists a normal nontrivial subgroup $ N\triangleleft G$.

And if I can show that the commutator subgroup $G'=1$ then since $G'=G$ I get a contradiction to the nontriviality assumption of the group.

Is this a good way to go? How should I view it? Thanks