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For endomorphisms and of a group G, define + by
( + )(x) =(x)(x).
Note: + need not be an endomorphism.
Lemma 6.33. If and are normal nilpotent endomorphisms of a finite indecomposable
group, and + is an endomorphism, then + is a normal nilpotent endomorphism.
Proof. It is obvious that + is normal. If it is an automorphism, then there exists a
such that ( + ) % = id. Set = and = . Then + = id, i.e.,
(x-1) (x-1) =x-1 =! (x) (x) =x = (x) (x) =! = .
Hence + = + . Therefore the subring of End(G) generated by and is commu-
tative. Because and are nilpotent, so also are and . Hence
m
( + )m = m + m-1 + + m
1
is zero for m sufficiently large.
GROUP THEORY 57
Proof. of Krull-Schmidt. Suppose G = G1 G2 Gs and G = H1 H2 Ht.
Write
i
i
! !
Gi ! G1 G2 Gs, Hi ! H1 H2 Ht.
i i
Consider 1 11 + 1 21 + =idG . Not all terms in the sum are nilpotent, and so,
1 2 1
after possibly renumbering the groups, we may suppose that the first is an automorphism,
say = 1 11 = -1. Thus (omitting subscripts)
1
(G1 ! G1 ! G ! H1 ! G ! G1) =idG .
1
Consider
(H1 ! G ! G1 ! G1 ! G ! H1) =.
Check % = (use above factorization of idG ), and so = id or 0. The second is impossible,
1
because occurs in idG % idG . Therefore, =idH . Hence 1 and 11 are isomorphisms.
1 1 1 1
On the other hand, 1(H2 ) = 1, but 11 =? is injective on G1. We conclude that
G1 )" (H2 ... Ht) =1. Henc e G1(H2 ) H" G1 (H2 ), and by counting, we see
that G = G1 H2 .
Repeat the argument.
Remark 6.34. (a) The Krull-Schmidt theorem holds also for an infinite group provided it
satisfies both chain conditions on subgroups, i.e., ascending and descending sequences of
subgroups of G become stationary. (See Rotman 6.33.)
(b) The Krull-Schmidt theorem also holds for groups with operators. For example, let
Aut(G) operate on G; then the subgroups in the statement of the theorem will all be char-
acteristic.
(c) When applied to a finite abelian group, the theorem shows that the groups Cm in
i
a decomposition G = Cm ... Cm are uniquely determined up to isomorphism (and
1 r
ordering).
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