$G$ 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:

$\{H\}=G_rH/H\leq G_{r-1}H/H\leq...\leq G_1H/H\leq G_0H/H=GH/H=G/H$

Now everything is well defined. What you have to do is show that each subgroup is normal in the next subgroup and that all quotients are abelian. No need to do that by definition, it is much better to use the correspondence theorem.


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