6.6: Krull-Schmidt theorem

A group \(G\) is indecomposable if \(G\neq1\) and \(G\) is not isomorphic to a direct product of two nontrivial groups, i.e., if

\[G\approx H\times H^{\prime}\implies H=1\text{ or }H^{\prime}=1. \nonumber \]

[ns28](a) A simple group is indecomposable, but an indecomposable group need not be simple: it may have a normal subgroup. For example, \(S_{3}\) is indecomposable but has \(C_{3}\) as a normal subgroup.

(b) A finite commutative group is indecomposable if and only if it is cyclic of prime-power order.

Of course, this is obvious from the classification, but it is not difficult to prove it directly. Let \(G\) be cyclic of order \(p^{n}\), and suppose that \(G\approx H\times H^{\prime}\). Then \(H\) and \(H^{\prime}\) must be \(p\)-groups, and they can’t both be killed by \(p^{m}\), \(m<n\). It follows that one must be cyclic of order \(p^{n}\), and that the other is trivial. Conversely, suppose that \(G\) is commutative and indecomposable. Since every finite commutative group is (obviously) a direct product of \(p\)-groups with \(p\) running over the primes, \(G\) is a \(p\)-group. If \(g\) is an element of \(G\) of highest order, one shows that \(\langle g\rangle\) is a direct factor of \(G\), \(G\approx\langle g\rangle\times H\), which is a contradiction.

(c) Every finite group can be written as a direct product of indecomposable groups (obviously).

[Krull-Schmidt][ns29]²⁵ Suppose that \(G\) is a direct product of indecomposable subgroups \(G_{1}% ,\ldots,G_{s}\) and of indecomposable subgroups \(H_{1},\ldots,H_{t}\):

\[G\simeq G_{1}\times\cdots\times G_{s},\quad G\simeq H_{1}\times\cdots\times H_{t}. \nonumber \]

Then \(s=t\), and there is a re-indexing such that \(G_{i}\approx H_{i}\). Moreover, given \(r\), we can arrange the numbering so that

\[G=G_{1}\times\cdots\times G_{r}\times H_{r+1}\times\cdots\times H_{t}. \nonumber \]

See , 6.36.

[ns30]Let \(G=\mathbb{F}_{p}\times\mathbb{F}_{p}\), and think of it as a two-dimensional vector space over \(\mathbb{F}_{p}\). Let

\[G_{1}=\langle(1,0)\rangle,\quad G_{2}=\langle(0,1)\rangle;\quad H_{1}% =\langle(1,1)\rangle,\quad H_{2}=\langle(1,-1)\rangle. \nonumber \]

Then \(G=G_{1}\times G_{2}\), \(G=H_{1}\times H_{2}\), \(G=G_{1}\times H_{2}\).

[ns31](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.

(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 characteristic.

(c) When applied to a finite abelian group, the theorem shows that the groups \(C_{m_{i}}\) in a decomposition \(G=C_{m_{1}}\times...\times C_{m_{r}}\) with each \(m_{i}\) a prime power are uniquely determined up to isomorphism (and ordering).