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I am reading Dummit and Foote section 3.4 (Composition series and the Holder program) and I am having trouble understanding how these concepts are connected.


A group $G$ is simple if the only normal subgroups are $\{1\}$ and $G$, i.e., $G$ has no non-trivial normal subgroups.

A subnormal series of groups $$1=N_0 \lhd N_1 \lhd \cdots \lhd N_k=G$$ is called a composition series if $N_{i+1}/N_i$ is simple for all possible $i$.

From this I gather that if a group $G$ is simple then its composition series must be trivial, i.e., $\{1\} \le G$.

A group $G$ is solvable if there is a subnormal series $$1=G_0 \lhd G_1 \lhd \cdots \lhd G_s=G$$ such that $G_{i+1}/G_i$ is abelian for all possible $i$.

From this I gather that every abelian group is solvable. Also, if a simple group is solvable, then it must be abelian.

How are these concepts further related?

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  • $\begingroup$ I think you have found the relations. But there will be some theorems, e.g., subgroups of solvable groups are solvable; quotient groups of solvable groups are solvable; if $N$ is normal in $G$, and if $N$ and $G/N$ are solvable, then $G$ is solvable. $\endgroup$ – Gerry Myerson Oct 23 '18 at 3:19
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    $\begingroup$ One minor correction: the trivial group is not considered to be a simple group. This means in particular that the subgroups in a composition series must all be distinct. $\endgroup$ – Derek Holt Oct 23 '18 at 7:51
  • $\begingroup$ What sort of answer are you expecting? For me this question is too broad (but I am always willing to be convinced otherwise!). $\endgroup$ – user1729 Oct 23 '18 at 20:33

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