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 $x^{-1}y^{-1}xy\in N$...

-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 $G/M$.

  • $\begingroup$ I can understand (1), (2) and (3). What do you mean by doing the same procedure for $G/M$? $\endgroup$
    – user441558
    Jun 4 '17 at 8:45
  • $\begingroup$ 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. $\endgroup$
    – KC.
    Jun 4 '17 at 9:02
  • $\begingroup$ $M$ 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? $\endgroup$
    – xunitc
    Sep 17 '20 at 14:47

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