Show that a finite group $G$ whose composition factors are of prime order implies having a normal series such that the factor groups are abelian

This appears as part of an exercise from Abstract Algebra: Dummit & Foote:

Let $$G$$ be a finite group:
(iii) all composition factors of $$G$$ are of prime order
(iv) $$G$$ has a chain of subgroups: $$1=N_0\unlhd N_1\unlhd N_2 \unlhd \dots \unlhd N_t=G$$ such that each $$N_i$$ is a normal subgroup of $$G$$ and $$N_{i+1}/N_i$$ is abelian, $$0\le i\le t-1$$
Prove that (iii) $$\Rightarrow$$ (iv).

Hint: ..., prove that a minimal nontrivial normal subgroup $$M$$ of $$G$$ is necessarily abelian and then use induction. To see that $$M$$ is abelian, let $$N\unlhd M$$ be of prime index and show that $$x^{-1}y^{-1}xy\in N$$ for all $$x,y\in M$$. Apply the same argument to $$gNg^{-1}$$ to show that $$x^{-1}y^{-1}xy$$ lies in the intersection of all $$G$$-conjugates of $$N$$, and use the minimality of $$M$$ to conclude that $$x^{-1}y^{-1}xy=1$$.

My attempt: I let $$M$$ be a minimal nontrivial normal subgroup of $$G$$. Such $$M$$ exists because $$G$$ is a nontrivial normal subgroup of itself. Among such subgroups there is a minimal one. Since $$M\unlhd G$$, there is a composition series of $$G$$, one of whose terms is $$M$$ (result of the previous exercise). Let $$N\unlhd M$$ be of prime index $$p$$. Let $$x,y\in M$$. Since $$|M/N|=p$$, $$M/N$$ is cyclic and its elements commute. So, $$xyN=yxN$$ and therefore $$x^{-1}y^{-1}xy\in N$$...

Questions:
-As $$M$$ is minimal nontrivial normal subgroup of $$G$$, why can't we immediately conclude that $$N=1$$?
-Even if I have to apply the same argument to $$gNg^{-1}$$, why is $$gNg^{-1}\unlhd M$$? $$m(gng^{-1})m^{-1}=(mgm^{-1})(mnm^{-1})(mg^{-1}m^{-1})=(mgm^{-1})n'(mg^{-1}m^{-1})=\dots$$
-And even if I know that $$x^{-1}y^{-1}xy$$ lies in the intersection of all $$G$$-conjugates of $$N$$, why is this intersection trivial, so that $$x^{-1}y^{-1}xy=1$$?
-Where do I use induction? I don't see any clue here.

As you can see, I'm pretty much hopeless solving this.

(1) $N$ being normal in $M$ does not imply that $N$ is normal in $G$.

(2) Try to show that $gNg^{-1}\trianglelefteq gMg^{-1}=M$.

(3) Prove that the intersection of $G$-conjugates of $N$ is normal in $G$. Then use minimality.

(4) Induction means that you do the same procedure for $G/M$.

• I can understand (1), (2) and (3). What do you mean by doing the same procedure for $G/M$?
– user441558
Commented Jun 4, 2017 at 8:45
• Suppose you have a minimal nontrivial normal subgroup of $G/M$. This normal subgroup is of the form $M_2/M$, where $M_2$ is normal in $G$ and contains $M$. By the same argument as before, $M_2/M$ is abelian. And so on until you find a normal series that must reach $G$ and each factor group is abelian.
– KC.
Commented Jun 4, 2017 at 9:02
• $M$ is abelian, the proof need composition factor $M/N$ is prime. Now say $M_2/M$ is abelian, how to proof all the composition factors of $G/M$ are prime please? Commented Sep 17, 2020 at 14:47