I recently read the well known theorem that for a group $G$ and $H$ a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. In my book, only the fact that $G$ is solvable implies $H$ is solvable was proven. I was able to show that if $H$ and $G/H$ are solvable, then so is $G$, but I can't quite show that $G$ is solvable implies $G/H$ is solvable.
My idea was this. Since $G$ is solvable, there exists a normal abelian tower $ G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}. $ I let $K_i=G_i/(H\cap G_i)$, in hopes of getting a sequence $ G/H\supset K_1\supset\cdots\supset K_r=\{e\}. $ My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that $K_{i+1}\unlhd K_i$ and that $K_i/K_{i+1}$ is abelian.
Writing $H_i=H\cap G_i$, I take $gH_{i+1}\in K_{i+1}$ for some $g\in G_{i+1}$. If g'H_i\in K_i, then I want to show g'H_igH_{i+1}g'^{-1}H_i is still in $K_{i+1}$, but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that $K_i/K_{i+1}$ is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you.