One can define the notion of "indecomposable" in many of the categories that mathematicians think about. However there's no real reason, as far as I can see, to expect it to behave at all well.

Here are two examples.

1) Say a group is indecomposable if it is not the trivial group, and not isomorphic to a product $H\times K$ with $H$ and $K$ groups, and neither of them the trivial group.

Can one find a group which can be written both as a product of two indecomposable groups, and as a product of three indecomposable groups?

2) Say a topological space is indecomposable if it has more than one element, and is not isomorphic to a product $X\times Y$, with $X$ and $Y$ topological spaces both having more than one element.

Can one find a topological space which can be written as a product of two indecomposable spaces, and also as a product of three indecomposable spaces?

In both cases the notion of indecomposability seems a bit artificial, or at least not commonly used, so I suspect that one can find such funny examples in both cases. I am pretty sure, for example, that it's not hard to write down a number field whose ring of integers contains elements which are both the product of two irreducibles and three irreducibles, and numbers are a lot easier than either groups or topological spaces. I don't know explicit examples for either Q1 or Q2 though. Does anyone else? I was told by an algebraist that no example of a finite group as in Q1 above can exist, which already surprised me a little.

Finite products and finite coproducts coincide in the category of [edit: abelian] groups I guess, but one could also formulate an analogue of Q2 using disjoint unions and this seems much more close to the kind of question that people do actually think about, so we can safely ignore it here :-)

  • 2
    $\begingroup$ Isn't the coproduct in the category of groups the free product? $\endgroup$ Commented Nov 10, 2011 at 0:45
  • $\begingroup$ A trivial example of a space that can be written as a product of 2 or 3 (pr generalize to any m,n) spaces --up to homeomorphism --is that of $\mathbb R^6$ , which can be written either as the product of 2 copies of $\mathbb R^3$ , or as the product of 3 copies of $\mathbb R^2$. This, of course, generalizes to using copies of any topological space X. $\endgroup$
    – gary
    Commented Nov 10, 2011 at 1:36
  • $\begingroup$ @gary: $\mathbb{R}^2$ is decomposable. $\endgroup$ Commented Nov 10, 2011 at 1:42
  • $\begingroup$ @Damian: you're right, I missed the big word here: indecomposable; good catch. $\endgroup$
    – gary
    Commented Nov 10, 2011 at 1:55
  • 1
    $\begingroup$ The list of groups which don't work for Q1 is much bigger, of course, via the Krull-Schmidt theorem. $\endgroup$
    – user641
    Commented Nov 10, 2011 at 3:14

1 Answer 1


There is a paper of A.L.S. Corner, A note on rank and direct decompositions of torsion-free Abelian groups, Proc. Cambridge Philos. Soc. 57 (1961) 230–233. (1961) where he shows the following. (I quote from the review on mathscinet.)

“Let $N$ and $k$ be natural numbers with $N>k$, and let $N=r_1+\cdots+r_k$ be any representation of the number $N$ as the sum of $k$ natural numbers. Then there exist an abelian group $G$ without torsion and subgroups $A_1,A_2,\ldots,A_k$ of $G$ such that

(a) the rank of subgroup $A_i$ is equal to $r_i$,
(b) $G=\sum_{i=1}^k A_i$ and
(c) the subgroups $A_i$ are indecomposable ($i=1,2,\ldots,k$).”

I haven't read this paper so I don't know whether the notation $\sum$ in this review means ‘direct sum’. If it does — and the Wikipedia article implies that it does — then this gives a positive answer to (1).

  • $\begingroup$ Shane: I found your answer incomprehensible until I went to the Wikipedia page and realised that the assertion proved in the paper is that, although $G$ depends on $N$ and $k$, it does not depend on the $r_i$! $\endgroup$ Commented Nov 10, 2011 at 8:11
  • $\begingroup$ @Kevin: Fair point. Curiously, I interpreted the result to mean that $G$ does not depend on the $r_i$, though this interpretation is not supported by the statement I quoted. $\endgroup$ Commented Nov 10, 2011 at 10:11
  • $\begingroup$ In fact, now I understand what your answer means, I note that it still doesn't actually answer the question I posed, because even though $G$ doesn't depend on the $r_i$, it does depend on $k$. However, the paper itself (whose review you quoted) contains an explicit example of (1). The group is abelian and a subgroup of $\mathbf{Q}^4$. $\endgroup$ Commented Nov 10, 2011 at 20:03

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