Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In this question, The Chaz asks whether $G\times G\cong H\times H$ implies that $G\cong H$, where $G$ and $H$ are finite abelian groups. The answer is to his question is yes, by the structure theorem for finite abelian groups, as noted in the answer by Anjan Gupta.

Even though I don't know the first thing about categories -- except for the things that I do know -- I'm wondering if and how such property could be expressed and proven in terms of universal properties, without actually manoevring inside the objects. For instance one may attempt to create a morphism $G\to H$ somehow appealing to the universal property of $\oplus$, and subsequently show this morphism is an isomorphism by chasing diagrams. But it seems likely that the existence of a structure theorem of some sort will be required.

This question may be considered trivial or weird for someone who's fluent with categories, I don't know. This topic contains quite a few references. I haven't really worked through any of them (yet) but I couldn't find anything helpful at first sight.

-- edit, So a more precise title would have been "Under what conditions that can be expressed in a universal way does $G\oplus G\cong H\oplus H$ imply $G\cong H$ ?" but I don't like long titles.

-- edit2, By the comment by Alexei Averchenko, maybe It's more natural to ask this question with the product instead of the coproduct. An answer to my question with the 'real' product would be appreciated too, obviously.

share|improve this question
Why the coproduct? –  Alexei Averchenko Mar 18 '11 at 6:00
@Alexei: Well that may actually be a very good question to begin with... I guess any product would do, if it helps to understand the case of finite abelian groups. –  Myself Mar 18 '11 at 6:04
I mean, why not ask about product if that's what your example is about? :) –  Alexei Averchenko Mar 18 '11 at 6:11
It's a just remark, but as far as I know both notions coincide in the category of abelian groups. The problem is that it feel it's not true for groups in general with the product (even though I'm having trouble finding a counterexample right now) whereas it seems less insane when thinking about the free product. It's more a guess, really. –  Myself Mar 18 '11 at 6:16
@Jonas: Yes I am aware of the fact. But this has everything to do with these object not being finitely generated which is a necessary condition for that structure theorem to exist. Because one way to answer my question would be to explain what such a structure theorem would look like in a categorical definition (therefore the existence of such theorem would exclude those objects from the category), I didn't really mention it. –  Myself Mar 18 '11 at 6:22
show 5 more comments

1 Answer

up vote 20 down vote accepted

Even in the category of all abelian groups, $G\oplus G\cong H\oplus H$ does not imply $G\cong H$: there are countably generated torsion free abelian groups $G$ such that $G\not\cong G\oplus G$, but $G\cong G\oplus G\oplus G$. Setting $H=G\oplus G$ gives an example where the implication does not hold. (The first examples were constructed by A.L.S. Corner, On a conjecture of Pierce concerning direct decompositions of abelian groups, Proc. Colloq. Abelian Groups (Tihany, 1963), Akadémiai Kiadó, Budapest, 1964, pp. 43-48; he proves that for any positive integer $r$ there exists a countable torsion free abelian group $G$ such that the direct sum of $m$ copies of $G$ is isomorphic to the direct sum of $n$ copies of $G$ if and only if $m\equiv n\pmod{r}$.)

Of course, the analogous result for direct products fails for groups, though it holds for groups having both chain conditions (by the Krull-Schmidt theorem).

The fact that you have categories with arbitrary coproducts/products (like the category of all groups and the category of all abelian groups) in which the result fails means that no proof via universal properties alone can exist. A proof that depended only the universal properties would translate into any category in which products/coproducts exist, but the result is false in general even for very nice categories.

Whether the conditions for the implication to hold can be expressed via universal conditions also seems to me to be doubtful. Universal conditions don't lend themselves easily to statements of the form "If something happens for $A$ and $B$, then it happens for $f(A)$ and $f(B)$" (where by "$f(-)$" I just mean "this other object cooked up or related to $A$ in some way") unless the construction/relation happens to be categorical.

share|improve this answer
Can you give a reference (or explanation) for how to construct the $G$ for which $G\not\cong G\oplus G$ but $G\cong G\oplus G\oplus G$? –  Anton Geraschenko Mar 18 '11 at 17:37
@Anton: A reference, at least; I've added it. –  Arturo Magidin Mar 18 '11 at 18:29
Thanks. Googling around, it looks like the OP's question is one of "Kaplansky's test problems" which people like to ask in various categories. Another one is, "If $G$ and $H$ are direct summands of each other, are they isomorphic?" The $G$ produced in Corner's paper obviously answers this problem in the negative for abelian groups. –  Anton Geraschenko Mar 18 '11 at 20:08
@Anton: Yes, there's a "family" of related questions. Also, "does $G\oplus H\cong G\oplus K$ imply $H\cong K$?" –  Arturo Magidin Mar 18 '11 at 20:12
@Arturo: thanks for your answer. But of course it fails for all abelian groups, one can just rip $\bigoplus_{i\in\mathbb Z} \mathbb Z$ into pieces, as was remarked by Jonas Meyer in a deleted comment. (Your example is a lot funnier, admitted.) And to the cancellation problem, consider $A\oplus B\oplus A\oplus B\oplus \dots$, which is isomorphic to itself, itself +A and itself +B, also see topic in the comment by Chandru1 beneath the question. However, your last paragraph is what really interests me: what is the $f$ that you have? And what is the precise meaning of 'not easily'? –  Myself Mar 19 '11 at 1:13
show 5 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.