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. ◻
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. ◻
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 , . ◻
Proof. . Since is indecomposable, either or . The latter implies is nilpotent. If , then and is a monomorphism, which implies that is an automorphism. ◻
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 is nilpotent. For large enough and all , . But this contradicts that and . ◻
Proof. Let
be the statement
. For
, let
be the statement: there is a reindexing of
such that
for
and
. We shall show inductively that
is true for all
such that
.
is true by hypothesis. Assume
is true.
for all
and
. Let
be the canonical epimorphisms to
. Similarly for
to
. Let
be the inclusion map sending
to
. Let
and let
. Verify the following identities:
Thus
for all
. The identities show that
Every sum of distinct
is a normal endomorphism. Since
is a normal automorphism of
and
satisfies both ACC and DCC on normal subgroups, for
some
,
,
is an automorphism on
.
is an automorphism of
. Since
and
,
cannot be nilpotent. Since
satisfies both chain conditions,
must be an automorphism of
. Therefore
is an isomorphism and so is
. To see this, note that
is equivalent to
.
is equivalent to
. Reindex the
so that we may assume
and
. Since
by the induction hypothesis,
is an internal direct product. For
and for
. Thus
Since
is an isomorphism,
. Thus
is an internal direct product. Define
as follows. Every
can be written as
. Let
. Clearly
.
is a monomorphism that is normal. Thus
is an automorphism so
. This proves
and completes the inductive argument. Therefore,
after reindexing,
for
. If
and if
,
. Since
are not trivial groups for all
, we must have
in either case. ◻