To discuss the Krull-Schmidt theorem, we must first define the ascending chain condition and descending chain condition.

*Proof.* Suppose
is not a finite
direct product of indecomposable subgroups. Let
be the set of all
normal subgroups
of
such that
is a direct factor of
and
is not a finite
direct product of indecomposable subgroups. Clearly
. If
, then
is not
indecomposable, whence there exist proper subgroups
and
of
such that
.
Furthermore, one of these groups, say
, must lie in
. Let
defined by
. There
exists a function
such that
and
.
Denoting
by
, we have a sequence
of normal subgroups
of
such that
.
If
satisfies the DCC on
normal subgroups, this is a contradiction. A routine inductive argument
shows
,
with each
a proper
subgroup of
. Thus there is a
properly ascending chain of normal subgroups:
If
satisfies the ACC on
normal subgroups, this is a contradiction.Â â—»

While it was easy to prove the existence of such a decomposition, proving that such a decomposition is unique turns out to be much more challenging. Unlike in the existence case, proving uniqueness requires both ACC and DCC on normal subgroups to hold.

*Proof.* Suppose
satisfies the ACC and
is an epimorphism.
The ascending chain of normal subgroups
must eventually become constant, say at
. Since
is an epimorphism, so
is
. If
for some
and
.
so
. Thus
is a monomorphism.
Next suppose
satisfies the DCC and
is a monomorphism.
is normal since
is a normal
endomorphism. The descending chain of normal subgroups
must become constant, say at
.
for some
. Since
is a monomorphism, so
is
and hence
.
Thus
is an
epimorphism.Â â—»

*Proof.* Consider the two chains of normal subgroups:
By hypothesis there is an
such that
for all
. Suppose
.
Then
for some
.
Thus
so
. Thus
.
so
for
some
.
thus
.
Since
,
.Â â—»

*Proof.*
.
Since
is indecomposable,
either
or
.
The latter implies
is nilpotent. If
,
then
and
is a monomorphism,
which implies that
is an
automorphism.Â â—»

We define some unconventional notation: if is a group and are functions, let be defined by . With given by , is a group under .

*Proof.* Since each
is a normal endomorphism, the proof will follow by induction once the
case is
established. If
is not
nilpotent, it is an automorphism. Note that the inverse
of
is a normal
automorphism. If
,
then
and
.
Thus
and
.
Thus
and
implying
.
An inductive argument shows that
.
Since each
is nilpotent,
has a
nontrivial kernel, whence