For $G$ a group and $H\unlhd G$, then $G$ is solvable iff $H$ and $G/H$ are solvable?

Question

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.

Answer 1

You're approach looks good to me. It might help to rewrite $K_i = G_i/(H\cap G_i)$ in the form $K_i = (G_i H)/H,$ so that you have a sequence of groups $G_i H$ all containing $H$, whose quotients are giving the $K_i$. Now to verify that $K_{i+1}$ is normal in $K_i$, you just have to verify that $G_{i+1} H$ is normal in $G_i H$, which isn't too hard to do. (Essentially, the computations with cosets that were giving you trouble have all been wrapped up once and for all into the isomorphism $G_i/(H\cap G_i) \cong (G_i H)/H$, and the coset computations with $G_{i+1} H$ inside $G_i H$ will be quite a bit easier.)

To see that $K_i/K_{i+1}$ is abelian, you can again use isomorphism theorems, to rewrite it as $(G_i H)/(G_{i+1} H) \cong G_i/(G_{i+1} H \cap G_i).$ You should be able to see that the latter group is a quotient of $G_i/G_{i+1}$, and so, being a quotient of an abelian group, is abelian.

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As a general remark, when studying the image of a subgroup $G'$ under a quotient map $G \to G/H$, passing back and forth between the descriptions $G'/G'\cap H$ and $G'H/H$ is a very standard method. The former description helps you think about the the image as a quotient of the given subgroup $G'$, while the latter description is useful for bringing into play the fact that "the lattice of subgroups of $G/H$ corresponds to the lattice of subgroups of $G$ containing $H$" --- it rewrites the image as the quotient of a subgroup containing $H$, and so helps you understand the inclusion relations and so on between different images as $G'$ varies.

Answer 2

I can't quite show that $G$ is solvable implies $G/H$ is solvable.

$G$ is solvable if and only if the process $G_0 = G, G_i = [G_{i-1}, G_{i-1}]$ of repeatedly taking commutator subgroups eventually terminates in the trivial group. If it terminates after $n$ steps, this is equivalent to saying that a certain word made of $n$ levels of nested commutators vanishes identically for every choice of elements of $G$, and this property is automatically preserved by homomorphisms $G \to G/H$ (as well as taking subgroups).