Condition for Composition Series

Theorem
Let $G$ be a finite group.

Then:
 * a normal series $\mathcal H$ for $G$ is a composition series for $G$


 * every factor group of $\mathcal H$ is a simple group.
 * every factor group of $\mathcal H$ is a simple group.

Proof
Let $G$ be a finite group whose identity is $e$.

Let:
 * $(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$

be a normal series for $G$.

Necessary Condition
Suppose there exists $k$ such that $G_{k + 1} / G_k$ is not a simple group.

Then there exists a normal subgroup $G'$ such that:
 * $\set e \lhd G' \lhd G_{k + 1} / G_k$

It follows that:
 * $G_k \lhd G'' \lhd G_{k + 1}$

where:
 * $G''$ is a normal subgroup of $G_{k + 1}$

and:
 * $G' = G'' / G_{k + 1}$

Thus $(1)$ has a proper refinement and so is not a composition series.

By the Rule of Transposition it follows that if normal series is a composition series, every factor group of that normal series is a simple group.

Sufficient Condition
Suppose $(1)$ is not a composition series for $G$.

Then a proper refinement of $(1)$ can be constructed by inserting a group $G''$ into the series somewhere, for example:
 * $G_k \lhd G'' \lhd G_{k + 1}$

It follows that $G / G_k$ is a normal subgroup of $G_{k + 1} / G$.

Thus, by definition, $G_{k + 1} / G_k$ is not a simple group.

Thus it has been shown that if a normal series is not a composition series, then it contains at least one factor group which is not a simple group

By the Rule of Transposition it follows that if every factor group of a normal series is a simple group, then that normal series is a composition series.

Hence the result.