Understanding the structure of a group through its decomposition in normal subgroups I am confused with a basic fact in group theory.
''Given a finite group $G,$ one can find a sequence of subgroups $$G=H_0\lhd H_1\lhd H_2 \cdot \cdot  \cdot \cdot \lhd H_r = \{e\}$$ and such that $H_{k+1}/H_k$ is simple.''
Since one talks about a group in general terms, then the statement must hold also for a simple group,whereby the sequence will be trivial. But in the case $G$ is not simple, by definition it must be solvable and the statement must hold again.
But then we read another statement in algebra books:
''Given the finite group $G,$ it is solvable if one can find a sequence of subgroups $$G=H_0\lhd H_1\lhd H_2 \cdot \cdot  \cdot \cdot \lhd H_r = \{e\}$$ and such that $H_{k+1}/H_k$ is abelian.''


*

*To my understanding, since the second statement is more specific than the first one, one would conclude that the abelian groups $H_{k+1}/H_k$ as stated in the second statement are simple (in which case they will be isomorphic to cyclic groups of prime order), something that is not stated in books. At the other hand, since the first statement must be true also for solvable groups, we must conclude that the simple group $H_{k+1}/H_k$ must be abelian in the case of a solvable group.


Can somebody comment on my conclusions and say what is wrong or true. Which statement implies which one, or, are they equivalent in the case of solvable groups ?


*One also reads in algebra books (I quote here Lang's book) the following: ''such a sequence already gives information about $G$. To get a full knowledge of $G$, one would have to know how these factor groups are pieced together.''


Can somebody explain through an example, what knowledge of $G$ we get through such a sequence and how to proceed to piece the factor groups together.
Many thanks.
 A: The first statement you wrote, that a finite group has a composition series
$$ 1 \lhd G_1 \lhd G_2 \lhd \cdots \lhd G_n = G $$
with all factors simple: this is called the Jordan Holder theorem. This is true for ALL finite groups.
The second type of series, where all the factors are abelian, does not always exist. The group is called "solvable" if such a series exists. But not all finite groups are solvable. Maybe I am misunderstanding your post, but it seems like maybe you are confusing the two notions. They are saying two different things.
As an example, take $G = S_5$, the symmetric group of degree 5. Then a Jordan Holder series is given by
$$ 1 \lhd A_5 \lhd S_5 $$
Here, $A_5$ is the alternating group. The factors are $A_5/1 \cong A_5$ and $S_5/A_5 \cong \Bbb{Z}_2$, which are both simple. However, $A_5$ is NOT abelian, so this is not an example of a solvable series. This illustrates my point that they are two distinct notions. This addresses your point #1 in your post.
To address your question/point #2, let's look at a smaller example: the symmetric group $S_3$. We have the same Jordan-Holder series as above:
$$ 1 \lhd A_3 \lhd S_3 $$
In this smaller case, $A_3 \cong \Bbb{Z}_3$, and this actually is also a solvable series. This is just a coincidence, since the example above shows this does not always happen. In any case, the factor groups are $\Bbb{Z}_3/1 \cong \Bbb{Z}_3$ and $S_3/\Bbb{Z}_3 \cong \Bbb{Z}_2$. Since $A_3 = \Bbb{Z}_3$ is a normal subgroup of $S_3$, and since we can identify the quotient $\Bbb{Z}_2$ with a subgroup of $S_3$ (say generated by one of the transpositions), then you can see that $S_3$ can be realized as the semidirect product $S_3 \cong \Bbb{Z}_3 \rtimes \Bbb{Z}_2$ (I'm leaving out some details...). So the group structure of $S_3$ is determined by the pieces and how they "fit together" (meaning the semidirect product structure).
