# The abelian tower of solvable quotient group $G/H$

$$G$$ is a solvable group, and so is $$H\lhd G$$. i.e., we have

$$\{e\}= G_r\lhd\dots\lhd G_1\lhd G_0=G$$

with $$G_i/G_{i+1}$$ abelian,

and $$H_i:=H\cap G_i$$ so that

$$\{e\}= H_r\lhd\dots\lhd H_1\lhd H_0=H$$

with $$H_i/H_{i+1}$$ abelian.

Now I want to show that $$G/H$$ is solvable by constructing an abelian tower for it. And what I construct was

$$\{H\}=G_r/H\lhd\dots \lhd G_1/H\lhd G_0/H=G/H,$$

but if $$H\not=\{e\}$$ we must have some $$j$$ such that $$|G_j|<|H|$$ provided that $$|G|<\infty$$ and it makes $$0<|G_j/H|<1$$, which is absurd.

If I quotient $$G_i$$ by $$H_i$$, then $$G_r/H_r=\{e\}/\{e\}=\{\{e\}\}=\{H_r\}\not=\{H\}$$, which is the trivial group of $$G/H$$.

Any hint or advice will be appreciated.

You are right. The problem is that $$H$$ might not be a subgroup of all groups in your normal sequence. So define the following sequence instead:
$$\{H\}=G_rH/H\leq G_{r-1}H/H\leq...\leq G_1H/H\leq G_0H/H=GH/H=G/H$$