# Abstract Algebra/Composition series

**Definitions 2.7.1**:

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

is simple.

**Theorem 2.7.2**:

Let be a finite group. Then there exists a composition series of .

**Proof**:

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

, then

, 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.

**Definition 2.7.3**:

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.

**Proof**:

**Theorem 2.7.5 (Jordan-Hölder)**:

Let be a group and let and be two composition series of . Then and are equivalent.

**Proof**:

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

and

, 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 :