Condition for Composition Series
Theorem
Let $G$ be a finite group.
Then:
- a normal series $\HH$ for $G$ is a composition series for $G$
- every factor group of $\HH$ 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:
- $G' = G / G_{k + 1}$
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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.
$\Box$
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 $G_{k + 1} / G$.
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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.
$\Box$
Hence the result.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Normal and Composition Series: $\S 74$
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $1$: Introduction to Finite Group Theory: $1.10$