- #1

Mr Davis 97

- 1,462

- 44

## 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.

## Homework Equations

## 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##. QEDHere 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##?