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
:
