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 $N$.

  • $\begingroup$ 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 $N$? $\endgroup$
    – graeme
    Aug 27 '14 at 23:43
  • 2
    $\begingroup$ @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 $G_0 N \leq G_1 N \leq \cdots G_n N$? $\endgroup$
    – user169852
    Aug 27 '14 at 23:54
  • $\begingroup$ of course, that is so obvious! Thank for your help! $\endgroup$
    – graeme
    Aug 27 '14 at 23:56
  • $\begingroup$ But i think it may actually work, after all the projection to the quotient group conserve the contention and is suprayective so maps normal subgroups in normal subgroups, even if it is not the so famous biyection $\endgroup$ Oct 14 '18 at 1:58

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