Let be a group. A normal series of are finitely many subgroups of such that
Two normal series and of are equivalent if and only if and there exists a bijective function such that for all :
A normal series of is a composition series of if and only if for each the group
Let be a finite group. Then there exists a composition series of .
We prove the theorem by induction over .
1. . In this case, is the trivial group, and with is a composition series of .
2. Assume the theorem is true for all , .
Since the trivial subgroup is a normal subgroup of , the set of proper normal subgroups of is not empty. Therefore, we may choose a proper normal subgroup of maximum cardinality. This must also be a maximal proper normal subgroup, since any group in which it is contained must have at least equal cardinality, and thus, if is normal such that
, which is why is not a proper normal subgroup of maximal cardinality.
Due to theorem 2.6.?, is simple. Further, since , the induction hypothesis implies that there exists a composition series of , which we shall denote by , where
. But then we have
, and further for each :
- is simple.
Thus, is a composition sequence of .
Our next goal is to prove that given two normal sequences of a group, we can find two 'refinements' of these normal sequences which are equivalent. Let us first define what we mean by a refinement of a normal sequence.
Let be a group and let be a normal sequence of . A refinement of is a normal sequence such that
Theorem 2.7.4 (Schreier):
Let be a group and let , be two normal series of . Then there exist refinements of and of such that and are equivalent.
Theorem 2.7.5 (Jordan-Hölder):
Let be a group and let and be two composition series of . Then and are equivalent.
Due to theorem 2.6.?, all the elements of must be pairwise different, and the same holds for the elements of .
Due to theorem 2.7.4, there exist refinements of and of such that and are equivalent.
But these refinements satisfy
, since if this were not the case, we would obtain a contradiction to theorem 2.6.?.
We now choose a bijection such that for all :