# 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 $H \lhd G$ we can construct the normal series:

- $(A): \quad \left\{{e}\right\} \lhd H \lhd G$

By Finite Group has Composition Series, $(A)$ can be refined to a composition series for $G$:

- $(B): \quad \left\{{e}\right\} = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$

Suppose $G_k = H$.

Then we can construct the composition series:

- $(C): \quad \left\{{e}\right\} = G_0 \lhd G_1 \lhd \cdots \lhd G_k = H$

and:

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

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.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 75$