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Could somebody please help me prove the following two statements:

  1. Any subgroup $H$ of a soluble group is itself soluble.
  2. If $G$ is soluble and $N$ is normal in $G$ then $G/N$ is soluble.

  1. I know we can consider the subnormal series of $G$ intersect $H$, but then why is each factor normal in the last and abelian?
  2. No idea

Thanks for any help!

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I'd like to reffer you to see this book: A Course on Finite Groups by H.E. Rose. You can find there a complete elementry facts about soluble groups. What you noted here are two Theorems at the begining of the chapter 11. –  Babak S. May 22 '12 at 16:49
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Your title talks about the "derived series", but your only progress expressed talks about using a subnormal series. In my answer I used the definition of solvable in terms of the existence of a subnormal series with abelian quotient; and you never mentioned derived series again. –  Arturo Magidin May 23 '12 at 0:16

2 Answers 2

up vote 3 down vote accepted
  1. Let $1=G_0\lhd G_1\lhd G_2\lhd\cdots\lhd G_n=G$ be a subnormal series of $G$ such that $G_{i+1}/G_i$ is abelian for each $i$, and let $H$ be a subgroup of $G$. Consider $$1 = G_0\cap H \subseteq G_1\cap H\subseteq \cdots \subseteq G_n\cap H = H.$$ Note that $G_i\lhd G_{i+1}$, hence $G_i\cap H\lhd G_{i+1}\cap H$. Thus, this is a subnormal series. Finally, we need to show that the quotients are abelian. The map $G_{i+1}\cap H\to G_{i+1}/G_i$ maps $G_{i+1}\cap H$ into an abelian group, hence the image is abelian; the kernel is precisely $G_i\cap H$, so $(G_{i+1}\cap H)/(G_i\cap H)$ is isomorphic to a subgroup of $G_{i+1}/G_i$, hence is abelian itself. Thus, $H$ is solvable.

  2. Consider the normal series $$1 = \frac{G_0N}{N} \lhd \frac{G_1N}{N} \lhd\cdots \lhd \frac{G_nN}{N} = \frac{G}{N}.$$ To show that $G_iN/N$ is normal in $G_{i+1}/N$ it suffices to show that $G_iN$ is normal in $G_{i+1}N$, since the lattice isomorphism theorem guarantees that normality is preserved in quotients. But since $G_i\triangleleft G_{i+1}$, and $N$ is normal in $G$, it follows that $G_{i}N\triangleleft G_{i+1}N$; indeed, if $gn\in G_{i}N$ and $xm\in G_{i+1}N$ with $x\in G_{i+1}$, then $$(xm)^{-1}gn(xm) = m^{-1}(x^{-1}gnx)m = m^{-1}(x^{-1}gx)(x^{-1}nx)m = (x^{-1}gx)n'$$ where $n'\in N$: $x^{-1}nx\in N$ since $N$ is normal, and for every $y\in G$ we have $Ny=yN$, so $m{-1}(x^{-1}gx) = (x^{-1}gx)m'$ for some $m'$. Now simply note that $x^{-1}gx\in G_i$ since $g\in G_i$, $x\in G_{i+1}$, and $G_i\lhd G_{i+1}$.

    Now, $$\frac{G_{i+1}N/N}{G_{i}N/N} \cong \frac{G_{i+1}N}{G_iN}.$$ I claim this is a quotient of $G_{i+1}/G_i$. Indeed, we can map $G_{i+1}$ to $G_{i+1}N/G_iN$ by mapping $g$ to $gG_iN$. The kernel contains $G_i$. So it is enough to show it is onto. Given $gn\in G_{i+1}N$, note that $gn\in gG_iN$, hence $gG_iN = gnG_iN$, so the map is indeed onto, and so $G_{i+1}N/G_iN$ is a quotient of $G_{i+1}/G_i$, hence is abelian.

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Amazing, Thanks! –  rk101 May 22 '12 at 17:22
    
Would you mind explaining sub-normality in (2) in further detail? –  rk101 May 22 '12 at 23:49
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@rk101: Don't accept an answer if you don't fully understand it! –  Arturo Magidin May 22 '12 at 23:49
    
Ok, Thanks for your help! I think I have got it all now. –  rk101 May 23 '12 at 0:06

Another way to approach this is to use an equivalent definition of a solvable (=soluble) group, namely that the derived series, i.e. the iterated series of commutator subgroups eventually reaches the trivial subgroup $1$ of $G$, that is
$G^{(0)}= G \trianglerighteq G^{(1)} \trianglerighteq G^{(2)} \trianglerighteq G^{(3)} ... \trianglerighteq G^{(k)} = 1$, for some $k \geqslant 0$, where $G^{(1)}= [G,G]$ and $G^{(i+1)} = [G^{(i)},G^{(i)}]$ for $i \geqslant 1$.
Now 1. follows from the fact that if $H \leqslant G$, then $H^{(i)} \leqslant G^{(i)}$. And 2. is implied by the fact that $(G/N)^{(i)} = G^{(i)}N/N$.

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Why is the derived series of $H$ a subgroup of the derived series of $G$? –  rk101 May 22 '12 at 22:01
    
The derived series of H is included in that of G. The individual terms are subgroups. This is by definition: the commutator subgroup $[H,H]$ of $H$ consists of the group generated by all $[h_1,h_2]$, with $h_1, h_2 \in H$, which trivially is a subgroup of $[G,G]$. The rest follows by induction. Clear now? –  Nicky Hekster May 22 '12 at 22:51
    
Yes, Thanks Nicky! –  rk101 May 23 '12 at 0:06
    
@rk101, you are welcome! Can you now prove the converse: if $N$ and $G/N$ are both solvable, then $G$ is solvable? –  Nicky Hekster May 25 '12 at 9:25

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