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Showing that every finite group has a composition series

  • #1
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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##. 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##?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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?
 
  • #4
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9,218
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\,.##
 

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