# Solvable implies quotient group is solvable: Proof check.

I'd like to check the veracity of my proof. I've seen several proofs using different methods (some I'm allowed to use with lots of element-pushing and others using ideas I'm not allowed), but none explicitly like mine.

First, a definition:

Definition: A group $G$ is solvable if there is a chain of subgroups $$1 = G_0 \triangleleft \dots \triangleleft G_k =G$$ such that $G_{i+1}/G_{i}$ is abelian for $i = 0, 1, \dots, k-1.$

Problem: Quotient groups of a solvable group are solvable.

Proof: Let $\overline{G}$ be the quotient of a solvable group $G$ by some normal subgroup $N$ of $G$. G is solvable so there is a chain:

$$1 = G_0 \triangleleft \dots \triangleleft G_n =G$$

such that $G_{i+1}/G_{i}$ is abelian for $i = 0, 1, \dots, n-1$. By the lattice isomorphism theorem, $$G_{i} \triangleleft G_{i+1} \iff \overline{G_{i}} \triangleleft \overline{G_{i+1}}$$ so there is a chain: $$\overline{1} = \overline{G_{0}} \triangleleft \dots \triangleleft \overline{G_{n}} = \overline{G}$$ for the quotient group. Now, by the 3rd isomorphism theorem, we have

$$\overline{G_{i+1}} / \overline{G_{i}} = (G_{i+1} / N)/(G_{i} / N) \cong G_{i+1} / G_{i}$$ which are abelian. $\Box$

The lattice isomorphism theorem gives you a correspondence between subgroups of $G/N$ and subgroups of $G$ which contain $N$. So your proof isn't quite right if any of the $G_i$ fail to contain $N$.
• Let me try again: The correspondence theorem works when the $G_{i}$ contain $N$. I also know that $N$ is the subgroup of a solvable group, so $N$ is solvable and has a composition series. Therefore I should be able to concatenate the two series once the $G_{i}$ no longer contain $N$? Aug 27 '14 at 23:43
• @user8476: How do you know that any of the $G_i$'s (besides $G$ itself) will contain $N$? What if you work with the subgroups $G_0 N \leq G_1 N \leq \cdots G_n N$?