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Demonstrate the following:

Let $G$ be a group and $N$ a normal subgroup. $G$ is solvable only if both $N$ and $G/N$ are solvable.

I'm working in Lang so am working with this characterization of solvability: A group $G$ is solvable only if it admits and abelian tower with the trivial group as the final element.

This is pretty basic; however I'm confusing myself when trying to show that $G$ slovable gives you $G/N$ solvable. My outline of an approach is as follows:

Let $G$ be solvable, then there is an abelian tower

$G=G_0 \subset G_1 \subset \ldots \subset G_n \subset \{e\}$

Then there is an abelian tower

$H=H_0 \subset H_1 \subset \ldots \subset H_n \subset \{e\}$

Where $H_i = G_i \cap H$. So H is solvable.

To show $G/H$ is solvable, my immediate reaction is to factor the $G_i$ by $H$, but that's not guaranteed to be well defined, so I'd want to look at the $G_i/H_i$. However, since consecutive $H_i$ are potentially different groups, possibly $G_{i+1}/H_{i+1} \subset G_i/H_i$ is not true, much less can we guarantee facorability.

Where am I confused?

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Before we go into too much detail, please allow me to ask, are you allowed to use the derived series definition of solvability? It makes for a much nicer proof. If $G$ is solvable and $\phi:G\twoheadrightarrow H$ then one can easily show that $\phi(G^{(i)})\supseteq H^{(i)}$ so that, if $G^{(n)}=\{e\}$ (which you know happens because $G$ is solvable) then $H^{(n)}=\{e\}$ and so $H$ is solvable. Your question is just then clearly a special case by taking $H=G/N$ and $\phi$ the canonical epimorphism. –  Alex Youcis Nov 1 '11 at 6:23
    
Aren't your set inclusions the wrong way around? $G$ is a subset of the one element group, the way you've written your tower. –  Gerry Myerson Nov 2 '11 at 1:52

1 Answer 1

Take the canonical map from $G$ to $G/N$, and consider its effect on the abelian tower for $G$.

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