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?