# Showing that every finite group has a composition series

## Homework Statement

Prove that for any finite group $G$ there exists a sequence of nested subgroups of $G$, $\{e\}=N_0\leq N_1\leq \cdots \leq N_n=G$ such that for each integer $i$ with $1\leq i\leq n$ we have $N_{i-1}\trianglelefteq N_i$ and the quotient group $N_i/N_{i-1}$ is simple.

## The Attempt at a Solution

Here is a proof I found online:

We prove the result by induction. Suppose that every group of order less than $|G|$ has a composition series.

Now if $G$ is a simple group then $\{ e_G \} \triangleleft G$ is a composition series. If $G$ is not a simple group then there exists a nontrivial proper normal subgroup of $G$. Since $G$ is a finite group, a maximal nontrovial proper normal subgroup exists. Denote this subgroup by $H$. Since $H$ is a proper subgroup of $G$ we have that $|H|<|G|$. By the induction hypothesis, $H$ has a composition series: $$\{e_G\}=H_0\le H_1\le\dots\le H_k=H.$$ But then $H \trianglelefteq G$ by the maximality of $H$. So: $$\{e_G\}=H_0\le H_1\le\dots\le H_k=H \le G.$$ The above chain of subgroup is a composition series of $G$. QED

Here are my questions: How do we know for sure that $G/H$ is simple? Is this implied by the maximality of $H$? Why?

Also, I am confused by the statement **but then $H \trianglelefteq G$ by the maximality of $H$.** Why does the maximality of $H$ matter in this case? Don't we already know that $H \trianglelefteq G$?

Related Calculus and Beyond Homework Help News on Phys.org
fresh_42
Mentor
Here are my questions: How do we know for sure that $G/H$ is simple? Is this implied by the maximality of $H$? Why?
If $G/H$ is not simple, then there is a normal subgroup $G/H \trianglerighteq K/H$ where $K\trianglelefteq G$ is a non-trivial normal subgroup which contains $H$.
Also, I am confused by the statement **but then $H \trianglelefteq G$ by the maximality of $H$.** Why does the maximality of $H$ matter in this case? Don't we already know that $H \trianglelefteq G$?
You are right. $H$ is chosen to be normal. Maximality is only needed for $G/H$ simple.

If $G/H$ is not simple, then there is a normal subgroup $G/H \trianglerighteq K/H$ where $K\trianglelefteq G$ is a non-trivial normal subgroup which contains $H$.

You are right. $H$ is chosen to be normal. Maximality is only needed for $G/H$ simple.
Could you explain this statement: Since $G$ is a finite group, a maximal nontrovial proper normal subgroup exists.

Why must a maximal nontrivial proper normal subgroup exist if $G$ is finite?

fresh_42
Mentor
Could you explain this statement: Since $G$ is a finite group, a maximal nontrovial proper normal subgroup exists.

Why must a maximal nontrivial proper normal subgroup exist if $G$ is finite?
We also have that $G$ is not simple, which is the real important property here. The case $G$ simple is trivial, so we assume that $G$ has a non-trivial normal subgroup. Among those we choose a maximal. The non-simplicity of $G$ gives us the subgroup we need, and the finiteness guarantees, that there is a maximal one, for otherwise we could have an infinite sequence $1 \triangleleft H_1\triangleleft H_2 \triangleleft \ldots G\,.$

• Mr Davis 97