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