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What is wrong with the following "proof" of the fact that a subgroup of a solvable group is solvable?

Definitions

Using the tower chain definition, we have that a group $G$ is solvable iff we can find a finite chain of subgroups $\{e\} \triangleleft N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_k = G$ such that each quotient group $N_{i+1}/ N_{i}$ is abelian. Note that $N \triangleleft G$ iff $\forall n \in N, \forall g \in G: gng^{-1} \in N$.

Lemma

Suppose $A \subseteq N \subseteq B$ are groups, $A \triangleleft B$, and $B/A$ is abelian. Then:

  • $A \triangleleft N$, since $A \triangleleft B \Rightarrow \forall a \in A, \forall b \in B: bab^{-1} \in A$ certainly implies that $\forall a \in A, \forall n \in N \subseteq B: nan^{-1} \in A \Rightarrow A \triangleleft N$.
  • $N/A$ is abelian, since $\forall b_1, b_2 \in B: b_1 b_2 A = b_2 b_1 A \Rightarrow \forall n_1, n_2 \in N \subseteq B: n_1 n_2 A = n_2 n_1 A$.

Proof

Suppose $G$ is solvable, and $N \subsetneq G$ is a non-trivial proper subgroup of $G$ (for $N = G$ or $N=\{ e \}$ the result is obvious). Let the series for $G$ be as in the definition. Then for some $i \in \{0, 1, \ldots, k \}$ we have that $N_i \subset N \subseteq N_{i+1}$. Then, by the lemma above, we have that $N_{i} \triangleleft N$ and $N/N_i$ is abelian. So we now have tower chain $\{e\} \triangleleft N_0 \triangleleft N_1 \triangleleft \ldots \triangleleft N_i \triangleleft N$ such that for $j \in \{1, \ldots, i \}$, each quotient group $N_{j}/ N_{j-1}$ and $N/N_i$ is abelian. So $N$ is solvable.

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  • $\begingroup$ Why would there be an $i$ such that $N_i \subset N \subseteq N_{i+1}$? $\endgroup$ Mar 31 '18 at 20:11
  • $\begingroup$ The chain "builds" up from the trivial group to the whole group, so at some point we must build $N$. $\endgroup$ Mar 31 '18 at 20:12
  • $\begingroup$ What you are saying is that there is an $i$ such that $N \subseteq N_{i+1}$, that is not enough. $\endgroup$ Mar 31 '18 at 20:15
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Unfortunately, the assumption that there must be an $i$ such that $N_i\subset N\subseteq N_{i+1}$ is false. As a counterexample, consider the symmetric group $G=S_4$ which is solvable by the chain $$\underset{=:N_0}{\underbrace{\{\mathrm{Id}\}}}\lhd\underset{=:N_1}{\underbrace{V_4}}\lhd \underset{=:N_2}{\underbrace{A_4}}\lhd \underset{=:N_3}{\underbrace{S_4}},$$ where $V_4$ is the Klein four group and $A_4$ the alternating group, and for example let $N:=\langle (1,2)\rangle $. As $(1,2)$ is an odd permutation, the only $i$ with $N\subseteq N_i$ is $N_3=S_4$. On the other hand, $N_j\not\subseteq N$ for all $0<j<3$ for obvious reasons.

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  • $\begingroup$ This is a good example; it shows a subgroup $N \subseteq G$ that exists "outside" the chain. $\endgroup$ Mar 31 '18 at 20:26

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