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This is Dummit & Foote $\S$3.4 #8:

Let $G$ be a finite group. Prove that the following are equivalent:

(i) $G$ is solvable

(ii) $G$ has a chain of subgroups: $1=H_0\trianglelefteq H_1 \trianglelefteq H_2\trianglelefteq\dots\trianglelefteq H_s = G$ such that $H_{i+1}/H_i$ is cyclic, $0\le i \le s-1$.

(iii) all composition factors of $G$ are of prime order

(iv) $G$ has a chain of subgroups: $1=N_0\trianglelefteq N_1 \trianglelefteq N_2 \trianglelefteq \dots \trianglelefteq N_t = G$ such that each $N_i$ is a normal subgroup of $G$ and $N_{i+1}/N$ is abelian, $0\le i\le t-1$

[For (iv), prove that a minimal nontrivial normal subgroup $M$ of $G$ is necessarily abelian, and then use induction. To see that $M$ is abelian, let $N\trianglelefteq M$ be of prime index (by (iii)) 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 $x^{-1}y^{-1}xy=1$.]

I have proved (i)$\Rightarrow$(iii)$\Rightarrow$(ii)$\Rightarrow$(i), and now I am trying to use the book's hint to prove that (iii)$\Rightarrow$(iv), but I'm confused as to why they even suggest taking $N\trianglelefteq M$. In my understanding, $M$ being a minimal normal subgroup indicates that there are no proper nontrivial subgroups contained in $M$. By an earlier exercise, I know that there is a composition series of $G$ which includes $M$, and since it is minimal it would have to take the following form: $$1\trianglelefteq M \trianglelefteq G_2 \trianglelefteq \dots\trianglelefteq G_r=G$$ Then we can use our hypothesis from (iii) to conclude that $M$ must have prime order, so $M$ is cyclic and therefore abelian. Given that $M$ is minimal, why do they even suggest considering some normal subgroup $N\trianglelefteq M$?

I don't think I'm really understanding the composition series concept correctly - for instance, in the book it mentions that the composition factors in a composition series are unique, that is if we have $1=N_0\le N_1 \le \dots \le N_r = G$ and $1=M_0 \le M_1 \le \dots \le M_s = G$ then $r=s$ and the composition factors are isomorphic (up to permutation), but I don't even agree with the $r=s$ part. Based on what I understand, $1\trianglelefteq \langle s\rangle\trianglelefteq \langle s,r^2\rangle\trianglelefteq D_8$ and $1\trianglelefteq\langle s\rangle\trianglelefteq \langle s\rangle \trianglelefteq \langle s, r^2 \rangle \trianglelefteq D_8$ are both valid composition series. This is really trivial, and to avoid it all that is required is to define composition series with proper normal subgroups, but that's not how it's defined in the text.

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The following claim in your question is wrong: "In my understanding, M being a minimal normal subgroup indicates that there are no proper nontrivial subgroups contained in M".

It must be: M being a minimal normal subgroup means there are no proper non-trivial normal subgroups (of the big group!) contained in M"

So what F&D are saying makes sense: there still can be some proper non-trivial subgroup N normal in M but not in G: $\,1<N\triangleleft M\,\,,\,\,N\rlap{\;/}\triangleleft G\,$

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Ah - that does make sense. I knew the normal part (forgot to type it in my question above), but forgot that it was in reference to the group as a whole ($G$). Thanks! –  process91 Sep 24 '12 at 16:15
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Yes, it is meant by proper normal subgroups, and uniqueness can be said only about maximal such composition series (having no more refinements).

The hint for (iii)$\Rightarrow$(iv) seems useless in your way, indeed, probably they thought about a proof like (i)$\Rightarrow$(iv)$\Rightarrow$(iii)$\Rightarrow$(ii).

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