Is it sufficient that G is solvable?

Question

Given a normal subgroup H of G, H and G/H are solvable. Then is G solvable? I know the converse is true... but I have no idea for our statement.

Answer 1

Yes, you are correct. For the proof, write down an abelian tower for $G/H$ starting at $1$ and ending at $G/H$. One can show that this lifts to an abelian tower starting at $H$ and ending at $G$. The abelian tower for $H$ completes the required tower.