The Krull–Schmidt theorem says:

If $G$ is a group that satisfies ACC and DCC on normal subgroups, then there is a unique way of writing $G$ as a direct product of finitely many indecomposable subgroups of $G$.

My question is:

If a group $G$ is not simple, doesn't this mean that $G$ is decomposable according to Krull–Schmidt Theorem?


The problem is that you are confusing having a normal proper normal subgroup with being decomposable. A group is decomposable if and only if it has a proper normal subgroup with a normal complement: that is, there have to exist normal subgroups $N$ and $M$, both proper, such that $N\cap M=\{1\}$ and $NM=G$. But in order to be non-simple, you only require the existence of a proper normal subgroup $N$.

Not every normal subgroup has a complement: this is not true even in abelian groups! Not every embedding $f\colon B\to A$ of a subgroup into an abelian group has a retraction (a homomorphism $g\colon A\to B$ such that $g\circ f = \mathrm{id}_B$), which is the condition required for $B$ to be a direct summand/direct factor of $A$.

This is exactly the same phenomenon you see when trying to diagonalize a linear transformation on a vector space of dimension $2$: in order for $T\colon\mathbf{V}\to\mathbf{V}$ to be diagonalizable you need to find two distinct one-dimensional subspaces that are $T$-invariant (proper $T$-invariant subspace play the role of normal subgroups above). For diagonalization to be impossible, one of two things must happen:

  1. There are no 1-dimensional $T$-invariant subspaces (this would be the parallel of the group being simple); this can happen over the real numbers, for example with a rotation by $90^{\circ}$.

  2. But it's possible for $T$ to not be diagonalizable, and yet for there to be a $1$-dimensional $T$-invariant subspace; but not two. An example is the linear transformation $T(x,y) = (x+y,y)$. The subspace $\{(x,0)\mid x\in F\}$ is $T$-invariant, but it's the only $T$-invariant subspace; there is no "second subspace" that you need to diagonalize. (This is the analogue of the cyclic group of order $p^2$, which has a unique proper nontrivial subgroup).

You seem to think that only the first case above can occur (no $T$-invariant subspace/ no normal subgroup). You are ignoring the possibility of the second case (there are $T$-invariant subspaces, just not enough of them/ there are normal subgroups, but none has a normal complement).


$\mathbb Z/p^2\mathbb Z$ is indecomposable but not simple.

  • $\begingroup$ so the theorem is not true for all finite groups? $\endgroup$ – Jawad Nov 25 '11 at 18:05
  • $\begingroup$ Simple group and indecomposable group are not equivalence. A simple group is indecomposable but an indecomposable group need not be simple, for example the group $\mathbb{Z}/p^{n}\mathbb{Z}$ is indecomposable but it is not simple. The Krull Schmidt theorem applies for the group of finite length(i.e length of the serie of normal subgroup in the factorization), it does not concern about the finiteness of the group. $\endgroup$ – Knumber10 Nov 25 '11 at 18:37
  • $\begingroup$ but all finite groups satisfy ACC and DCC on normal subgroups? $\endgroup$ – Jawad Nov 25 '11 at 20:54
  • $\begingroup$ @Jawad AFAICS, if G is indecomposable the decomposition from the theorem consists of one indecomposable subgroup, namely G itself $\endgroup$ – Grigory M Nov 25 '11 at 21:31
  • $\begingroup$ thank you all but it still fuzzy $\endgroup$ – Jawad Nov 25 '11 at 23:33

The best example of a finite indecomposable group which is not simple is $\mathbb{S}_n$ for $n\geq 3$ (Permutation group). This is easy to prove as well.


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