Let $H$ be a proper subgroup of maximal order in a group $G$. Now suppose $H$ is solvable and normal in $G$ so there exists a series of normal subgroups:
$1 = H_0 ⊲ H_1 ⊲ ... ⊲ H_n = H$, with $\dfrac{H_{j+1}}{H_j}$ abelian. Consider the cyclic group $<g>$ = $K$ generated by some $g\in G-H$. Now $HK = G$ by maximality of $H$. Thus $\dfrac{G}{H}\cong\dfrac{HK}{H}\cong\dfrac{K}{H∩K}$ wich is a quotient of an abelian group ($K$ is cyclic and hence abelian) so $G/H$ is abelian. We can now construct a series of normal subgroups as following: $1 = H_0 ⊲ H_1 ⊲ ... ⊲ H_n = H ⊲ G = H_{n+1}$ with $\dfrac{H_{j+1}}{H_j}$ abelian. So G is solvable.
My question is, is there another way to approach this proof or in some way generalize it?
Can we in some way have less restrictions on $H$ and still make the proof work for example?
