Group G is Solvable If and Only If H and G/H are Solvable

abstract-algebragroup-theory

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.

Best Answer

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.


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.