How to break permutation group in normal subgroup and the quotient group?

Question

could you give steps to show this decomposition

or show in maple code

http://en.wikipedia.org/wiki/Simple_group

Answer 1

Let $G$ be a finite group. The set of all non-identity normal subgroups of $G$ is a non-empty finite set (it's finite because $G$ is finite and it's nonempty because $G$ is a normal subgroup of itself). Therefore, there exists a minimal normal subgroup $N_1$ of $G$. $N_1$ need not be unique! (See the exercises below.)

The quotient $G/N_1$ (which exists because $N_1$ is normal in $G$) in turn possesses a minimal normal subgroup. Under the canonical bijection between subgroups of $G/N_1$ and subgroups of $G$ containing $N_1$, this minimal normal subgroup of $G/N_1$ corresponds to a normal subgroup of $G$ containing $N_1$. Let's denote it by $N_2$. $N_2$ is minimal with respect to the property that it is a normal subgroup of $G$ properly containing $N_1$. $N_2$ need not be unique! (... for the same reason that $N_1$ need not be unique!)

The process continues inductively and it must terminate because $G$ is finite; the last term of the process will, of course, be $G$. So, we end up with a sequence

$\{e\}=N_0\triangleleft N_1\triangleleft\cdots\triangleleft N_{k-1}\triangleleft N_k=G$

where the notation $\triangleleft$ signifies "is a normal subgroup of" and $e$ denotes the identity element of $G$. Such a series is known as a composition series of the group $G$. Note that each quotient group in this series $N_{i+1}/N_i$ for all $0\leq i\leq k-1$ is a simple group. The quotients are sometimes known as the composition factors of the composition series. The composition length of $G$ is the number of non-identity terms in a composition series of $G$ (so it's $k$, not $k+1$, in the notation of the process described above). It turns out that the composition length doesn't depend on which composition series of $G$ you choose. So we can speak of the composition length!

Exercise 1: Verify that $\{e\}\triangleleft \{(1 2),(1 3),(2 3)\}\triangleleft S_3$ and $\{e\}\triangleleft \{(1 2 3),(1 3 2)\}\triangleleft S_3$ are composition series for $S_3$.

Exercise 2: What is the composition length of the following groups:

(a) An abelian group?

(b) The permutation group $S_3$?

(c) (challenge) The permutation group $S_4$?

(d) (challenge) The alternating group $A_4$? (If you couldn't figure out (c), then what is the composition of length in terms of the (possibly unknown) composition length of $S_4$?)

Exercise 3: What are the composition factors of the groups you considered in Exercise 2?

Exercise 4 (Jordan-Holder): Prove that the composition length and the isomorphism classes of the composition factors of a finite group are independent of the chosen composition series. (This exercise is difficult and it's known as the Jordan-Holder theorem; try induction on the order of the group to get started.)

Exercise 5: Prove that there is a unique composition series of $S_n$ for all $n\geq 5$ and find it! (Hint: first prove that the only non-trivial proper normal subgroup of $S_n$ for $n\geq 5$ is $A_n$, the alternating group on $n$ letters.)

I hope this helps!