Group is Solvable iff Normal Subgroup and Quotient are Solvable

Theorem
Let $G$ be a finite group.

Let $H$ be a normal subgroup of $G$.

Then $G$ is solvable iff:
 * $(1): \quad H$ is solvable

and
 * $(2): \quad G / H$ is solvable

where $G / H$ is the quotient group of $G$ by $H$.

Proof
As $H \triangleleft G$ we can construct the normal series:
 * $(A): \quad \left\{{e}\right\} \triangleleft H \triangleleft G$

By Finite Group has Composition Series, $(A)$ can be refined to a composition series for $G$:
 * $(B): \quad \left\{{e}\right\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G$

Suppose $G_k = H$.

Then we can construct the composition series:
 * $(C): \quad \left\{{e}\right\} = G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_k = H$

and:
 * $(D): \quad \left\{{e}\right\} = G_k / H \triangleleft G_{k+1} / H \triangleleft \cdots \triangleleft G_n / H = G / H$

Furthermore, by the Third Isomorphism Theorem:


 * $\dfrac {G_{i+1} / H} {G_i / H} \cong \dfrac {G_{i+1}} {G_i}$

for all $k \le i \le n$.

So each factor of the composition series for $G$ is a factor of either:
 * the composition series for $H$

or:
 * the composition series for $G / H$.

The result follows.