8
$\begingroup$

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 :-)

  • 0
    Probably should also note, in the context of Shane's comment on the coproduct, that a finitely generated group $G$ has a unique decomposition as a free product of freely indecomposable groups - this is known as the Grushko decomposition theorem.2011-11-10

1 Answers 1

4

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).

  • 0
    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 $\mathbf{Q}^4$.2011-11-10