As an example, the derived series is a normal series for any group . If is nilpotent, the ascending central series is a normal series for .
A commonly used fact for composition series: When , is simple iff is maximal in the set of all normal subgroups of with .
Proof. (i) Let be a maximal normal subgroup of . Then is simple. Let be a maximal normal subgroup of and so on. Since is finite, this process must end with . Thus is a composition series.
(ii) If is abelian, , then is abelian since it is a subgroup of and is abelian since it is isomorphic to .
(iii) If , is a proper normal subgroup of . All proper normal subgroups of have the form for some . ◻
Proof. If is solvable, is a solvable series. If is a solvable series for , then abelian implies that . abelian implies . Proceeding by induction conclude . In particular, so is solvable. ◻
Proof. A composition series with cyclic factors is a solvable series. Conversely, assume is a solvable series for . If , let be a maximal normal subgroup of containing . If , let be a maximal normal subgroup of containing . Continue until we obtain a series with each subgroup a maximal normal subgroup of the preceding, whence each factor is simple. This series terminates as in, eventually since is finite. Doing this for each pair gives a solvable refinement of the original series. Each factor of this series is abelian and simple hence cyclic of prime order. ◻
Proof. Since is a composition series, has no proper refinements. Thus any refinement of is obtained by inserting additional copies of . Any refinement of has the exact same nontrivial factors as and is thus equivalent to . ◻
Proof. Note . Similarly, . Thus . Also, . We will define an epimorphism with kernel . This would imply that and that . Define as follows: If , let . implies . Thus is well defined. is clearly surjective. Finally, . Hence . . A symmetric argument shows (ii) and whence (iii) follows. ◻
Proof. Let and be subnormal [resp. normal] series. Let and for each , consider the groups For each , the Zassenhaus lemma applied to shows that (if the original series is normal, each ) Inserting these groups between each and , denoting by gives a subnormal (resp. normal) refinement of . A symmetric argument shows there is a refinement of with . By the Zassenhaus lemma, . ◻
Proof. Since composition series are subnormal series, any two composition series have equivalent refinements by the Schreier Refinement Theorem. But every refinement of a composition series is equivalent to the original composition series. Thus any two composition series are equivalent. ◻