a finite group is solvable ⇔ derived series reaches {e}

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Direction 1: If the derived series of $G$ reaches the trivial group, then $G$ is solvable
Suppose the derived series $G = G^{(0)} \triangleright G^{(1)} \triangleright G^{(2)} \triangleright \cdots \triangleright G^{(k)} = \{e\}$ for some finite $k$, where each $G^{(i)} = [G^{(i-1)}, G^{(i-1)}]$. This is a normal series because the commutator subgroup of any group is normal in that group, so each $G^{(i)}$ is normal in $G^{(i-1)}$ (and hence subnormal in $G$). Moreover, each factor group $G^{(i-1)} / G^{(i)}$ is abelian, since $G^{(i)}$ is precisely the commutator subgroup of $G^{(i-1)}$, and the quotient by the commutator subgroup is always abelian. Thus, $G$ admits a subnormal series with abelian factors, so $G$ is solvable by definition.
Direction 2: If $G$ is solvable, then its derived series reaches the trivial group
We proceed by induction on the order $|G|$.

• Base case: If $|G| = 1$, then $G$ is trivial, and its derived series is already $\{e\}$.
• Inductive hypothesis: Assume that for all solvable groups $H$ with $|H| < |G|$, the derived series of $H$ terminates at the trivial group.
• Inductive step: Let $G$ be a finite solvable group with $|G| > 1$. If $G$ is abelian, then $G^{(1)} = [G, G] = \{e\}$, so the derived series terminates immediately. Now assume $G$ is not abelian. The derived subgroup $G^{(1)} = [G, G]$ is a proper subgroup of $G$ because the quotient $G / G^{(1)}$ is abelian and non-trivial (since $G$ is not abelian). Thus, $|G^{(1)}| < |G|$. Subgroups of solvable groups are solvable, so $G^{(1)}$ is solvable. By the inductive hypothesis, the derived series of $G^{(1)}$ terminates at the trivial group: $G^{(1)} \triangleright G^{(2)} \triangleright \cdots \triangleright G^{(k+1)} = \{e\}$ for some $k$. But this is precisely the tail of the derived series of $G$, starting from $G^{(1)}$. Therefore, the full derived series of $G$ is $G \triangleright G^{(1)} \triangleright G^{(2)} \triangleright \cdots \triangleright G^{(k+1)} = \{e\}$, which terminates at the trivial group.

By induction, the result holds for all finite solvable groups.

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Let $G$ be a nilpotent group. By definition, its lower central series terminates at the trivial subgroup: $G = \gamma_1(G) \triangleright \gamma_2(G) \triangleright \cdots \triangleright \gamma_c(G) = \{e\}$ for some integer $c \geq 1$, where $\gamma_1(G) = G$ and $\gamma_{k+1}(G) = [G, \gamma_k(G)]$.
To show $G$ is solvable, we must prove that its derived series also terminates at $\{e\}$: $G = G^{(0)} \triangleright G^{(1)} \triangleright \cdots \triangleright G^{(m)} = \{e\}$ for some $m$, where $G^{(k)} = [G^{(k-1)}, G^{(k-1)}]$. We prove by induction on $k$ that $G^{(k)} \leq \gamma_{k+1}(G)$ for all $k \geq 0$.

• Base cases: For $k=0$, $G^{(0)} = G = \gamma_1(G)$. For $k=1$, $G^{(1)} = [G, G] = \gamma_2(G)$.
• Inductive hypothesis: Assume $G^{(k-1)} \leq \gamma_k(G)$ for some $k \geq 2$.
• Inductive step: Then $G^{(k)} = [G^{(k-1)}, G^{(k-1)}]$. Since $G^{(k-1)} \leq \gamma_k(G) \leq G$, we have $[G^{(k-1)}, G^{(k-1)}] \leq [G, G^{(k-1)}]$ (because commutators in the former are special cases of those in the latter, with the first element from $G^{(k-1)} \subseteq G$). Moreover, $[G, G^{(k-1)}] \leq [G, \gamma_k(G)] = \gamma_{k+1}(G)$ (because $G^{(k-1)} \leq \gamma_k(G)$, so the subgroup generated by commutators $[g, h]$ with $h \in G^{(k-1)}$ is contained in the subgroup generated by such with $h \in \gamma_k(G)$). Thus, $G^{(k)} \leq \gamma_{k+1}(G)$.

By induction, $G^{(k)} \leq \gamma_{k+1}(G)$ for all $k$.
Since $G$ is nilpotent, $\gamma_c(G) = \{e\}$, and the lower central series stabilizes: $\gamma_k(G) = \{e\}$ for all $k \geq c$. Choose $k \geq c-1$, so $k+1 \geq c$. Then $G^{(k)} \leq \gamma_{k+1}(G) = \{e\}$, hence $G^{(k)} = \{e\}$. Therefore, the derived series terminates at $\{e\}$, and $G$ is solvable.

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Solvability: The derived series $S_3 \triangleright A_3 \triangleright \{e\}$ terminates at {e}.
Not nilpotent: The lower central series $S_3 \triangleright A_3 \triangleright A_3 \triangleright \cdots$ does not terminate at {e}.

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Since the sequence is short exact, $A$ is a normal subgroup of $B$ and $C \cong B/A$.
A group is solvable if and only if its derived series terminates at the trivial subgroup: $G^{(k)} = \{e\}$ for some finite $k$, where $G^{(0)} = G$ and $G^{(i)} = [G^{(i-1)}, G^{(i-1)}]$.
Assume $A^{(m)} = \{e\}$ and $C^{(n)} = \{e\}$ for some finite $m, n$.
We show that $(B/A)^{(k)} = B^{(k)} A / A$ for all $k \geq 0$, by induction on $k$.

• Base case ($k=0$): $(B/A)^{(0)} = B/A = B A / A = B/A$ (since $A \leq B$).
• Inductive hypothesis: Assume $(B/A)^{(k)} = B^{(k)} A / A$.
• Inductive step: Let $H = B^{(k)}$. Then $(B/A)^{(k+1)} = [(B/A)^{(k)}, (B/A)^{(k)}] = [H A / A, H A / A] = [H A, H A] / A$. Now, $[H A, H A] \leq [H, H] [H, A] [A, A]$ (proof of this inclusion provided below). Since $[H, A] \leq A$ and $[A, A] \leq A$ (as $A$ is normal in $B$), it follows that $[H A, H A] \leq [H, H] A$. Conversely, $[H, H] \leq [H A, H A]$, so $[H A, H A] A = [H, H] A$. Thus, $[H A, H A] / A = [H, H] A / A = B^{(k+1)} A / A$.

By induction, $(B/A)^{(k)} = B^{(k)} A / A$ for all $k$.
Since $C^{(n)} = \{e\}$, we have $B^{(n)} A / A = \{e\}$, so $B^{(n)} \leq A$.
Now, $B^{(n+1)} = [B^{(n)}, B^{(n)}] = (B^{(n)})^{(1)} \leq A^{(1)}$, and inductively, $B^{(n+j)} = (B^{(n)})^{(j)} \leq A^{(j)}$ for all $j$ (since the derived series is decreasing and $B^{(n)} \leq A$). Thus, $B^{(n+m)} \leq A^{(m)} = \{e\}$, so $B^{(n+m)} = \{e\}$.
Therefore, $B$ is solvable.

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Detailed proof of the inclusion $[HA, HA] \leq [H,H][H,A][A,A]$
We adhere to the notation from Wikipedia’s commutator identities page: the conjugate of an element $g$ by $k$ is written $g^k = k^{-1} g k$.
The subgroup $[HA, HA]$ is generated by commutators of the form $[ha, h'a']$ for $h, h' \in H$ and $a, a' \in A$.
Compute $[ha, h'a'] = (ha)^{-1} (ha)^{h'a'}$, since $[x,y] = x^{-1} x^y$.
This equals $a^{-1} h^{-1} (ha)^{h'a'}$.
Now, $(ha)^{h'a'} = h^{h'a'} a^{h'a'}$, so $[ha, h'a'] = a^{-1} h^{-1} h^{h'a'} a^{h'a'} = a^{-1} [h, h'a'] a^{h'a'}$, since $h^{-1} h^{h'a'} = [h, h'a']$.
Using the identity $g^k = g [g, k]$, we have $a^{h'a'} = a [a, h'a']$, so the expression is $a^{-1} [h, h'a'] a [a, h'a'] = [h, h'a']^a [a, h'a']$.
Now expand $[h, h'a']$ using the identity $[x, yz] = [x, z] [x, y]^z$: here $h'a' = h' a'$, so $[h, h' a'] = [h, a'] [h, h']^{a'}$.
Thus, $[h, h'a']^a = ([h, a'] [h, h']^{a'})^a = [h, a']^a [h, h']^{a' a}$.
But since $A$ is normal in $B$, and $H = B^{(k)}$ is a characteristic subgroup of $B^{(k-1)}$ (hence normal in $B$), the commutator subgroups $[H,H]$, $[H,A]$, and $[A,A]$ are normal in $B$.
Therefore, conjugations by elements of $A$ (like $a$ or $a'$) map these subgroups to themselves: $[h, h']^{a' a} \in [H,H]$, $[h, a']^a \in [H,A]$.
So $[h, h'a']^a \in [H,A] [H,H]$.
Similarly, expand $[a, h'a'] = [a, h' a'] = [a, a'] [a, h']^{a'}$, using $[x, yz] = [x, z] [x, y]^z$.
So $[a, a'] \in [A,A]$, $[a, h']^{a'} \in [H,A]$.
Thus, $[a, h'a'] \in [A,A] [H,A]$.
Putting it together, $[ha, h'a'] \in [H,H] [H,A] [A,A]$.
Since this holds for all generators, and the product is a subgroup, $[HA, HA] \leq [H,H] [H,A] [A,A]$.

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Since the sequence is short exact, $A$ is a normal subgroup of $B$ and $C \cong B/A$. Let $\pi: B \to B/A$ be the canonical projection homomorphism.

A group is solvable if and only if its derived series terminates at the trivial subgroup: $G^{(k)} = \{e\}$ for some finite $k$, where $G^{(0)} = G$ and $G^{(i)} = [G^{(i-1)}, G^{(i-1)}]$.

Assume $A^{(m)} = \{e\}$ and $C^{(n)} = \{e\}$ for some finite $m, n$.

Since $\pi$ is a homomorphism, $\pi([x,y]) = [\pi(x), \pi(y)]$, so $\pi(B^{(1)}) = [B/A, B/A] = (B/A)^{(1)}$. By induction, $\pi(B^{(k)}) = (B/A)^{(k)}$ for all $k \geq 0$.

In particular, since $C^{(n)} = \{e\}$, we have $\pi(B^{(n)}) = (B/A)^{(n)} = \{e\}$, so $B^{(n)} \leq \ker \pi = A$.

Now, $B^{(n+1)} = [B^{(n)}, B^{(n)}] = (B^{(n)})^{(1)} \leq A^{(1)}$, and inductively, $B^{(n+j)} = (B^{(n)})^{(j)} \leq A^{(j)}$ for all $j$ (since the derived series is decreasing and $B^{(n)} \leq A$). Thus, $B^{(n+m)} \leq A^{(m)} = \{e\}$, so $B^{(n+m)} = \{e\}$.

Therefore, $B$ is solvable.