Krull–Schmidt theorem

In mathematics, the Krull–Schmidt theorem states that a group subjected to certain finiteness conditions on chains of subgroups, can be uniquely written as a finite direct product of indecomposable subgroups. We say that a group G satisfies the ascending chain condition (ACC) on subgroups if every sequence of subgroups of G: is eventually constant, i.e., there exists N such that GN = GN+1 = GN+2 = ... . We say that G satisfies the ACC on normal subgroups if every such sequence of normal subgroups of G eventually becomes constant. Likewise, one can define the descending chain condition on (normal) subgroups, by looking at all decreasing sequences of (normal) subgroups: Clearly, all finite groups satisfy both ACC and DCC on subgroups. The infinite cyclic group satisfies ACC but not DCC, since (2) > (2)2 > (2)3 > ... is an infinite decreasing sequence of subgroups. On the other hand, the -torsion part of (the quasicyclic p-group) satisfies DCC but not ACC. We say a group G is indecomposable if it cannot be written as a direct product of non-trivial subgroups G = H × K. If is a group that satisfies either ACC or DCC on normal subgroups, then there is exactly one way of writing as a direct product of finitely many indecomposable subgroups of . Here, uniqueness means direct decompositions into indecomposable subgroups have the exchange property. That is: suppose is another expression of as a product of indecomposable subgroups. Then and there is a reindexing of the 's satisfying and are isomorphic for each ; for each . Proving existence is relatively straightforward: let S be the set of all normal subgroups that can not be written as a product of indecomposable subgroups. Moreover, any indecomposable subgroup is (trivially) the one-term direct product of itself, hence decomposable. If Krull-Schmidt fails, then S contains G; so we may iteratively construct a descending series of direct factors; this contradicts the DCC. One can then invert the construction to show that all direct factors of G appear in this way.