Any Finite Group Has a Composition Series

Group Theory Problems and Solutions

Problem 122

Let $G$ be a finite group. Then show that $G$ has a composition series.

 
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Proof.

We prove the statement by induction on the order $|G|=n$ of the finite group.
When $n=1$, this is trivial.


Suppose that any finite group of order less than $n$ has a composition series.
Let $G$ be a finite group of order $n$.
If $G$ is simple, then $G \rhd \{e\}$, where $e$ is the identity element of $G$, is a composition series and we are done.

Thus, suppose that $G$ is not simple. Then it has a nontrivial proper normal subgroup.
Since $G$ is a finite group, there exists a maximal proper normal subgroup $H$.

Then the quotient $G/H$ is a simple group.
In fact, if $N$ is a proper normal subgroup of $G/H$, then the preimage of $N$ under the natural projection homomorphism $\pi:G \to G/H$ is a proper normal subgroup of $G$ containing $H$ by the fourth isomorphism theorem.
Since $H$ is maximal, the preimage $\pi^{-1}(N)$ must be $H$. This implies $N$ is trivial in $G/H$ and thus $G/H$ is simple.

Since $H$ is a proper subgroup of $G$, the order of $H$ is less than that of $G$.
Thus by the induction hypothesis, $H$ has a composition series
\[H=N_1 \rhd N_1 \rhd \cdots \rhd N_m=\{e\}.\]

The series
\[G=N_0 \rhd H=N_1 \rhd N_1 \rhd \cdots \rhd N_m=\{e\}\] has a simple factors $N_i/N_{i+i}$, hence it is a composition series for $G$.

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Infinite cyclic groups do not have composition series


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  1. 09/27/2016

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