Let $G$ be a retract of a product $\prod_I\mathbb{Z}$ of copies of $\mathbb Z$. This means there are group homomorphisms $\pi:\prod_I\mathbb{Z}\to G$ and $\sigma:G\to\prod_I\mathbb Z$ such that $\pi\circ\sigma=Id_G$. Is it true that $G$ must be a product of copies of $\mathbb Z$?

  • $\begingroup$ The introduction to R.J. Nunke’s On direct products of infinite cyclic groups (Proc. Amer. Math. Soc. 13 1962 (66-71)) says that the paper proves that the answer to the OP’s question is “yes”. However, on a quick look through the paper I don’t see where this is proved in full generality. Probably I’m missing something. $\endgroup$ Jun 11 at 11:15

1 Answer 1


EDIT: In an earlier version, I claimed erroneously that the following statement holds for general $I$, and can be formally reduced to the case of countable $I$. I currently do not see how to perform that reduction. The argument also had some gaps, which I hopefully fixed now. So here's an argument for countable $I$:

For countable $I$, we have $\operatorname{Hom}(\prod_I \mathbb{Z}, \mathbb{Z})\cong \bigoplus_I \mathbb{Z}$. The key statement is that a homomorphism $\prod_I \mathbb{Z} \to \mathbb{Z} $ necessarily has "finite support", i.e. there is a finite subset $I'\subseteq I$ such that $f$ factors through the projection $\prod_I \mathbb{Z}\to \prod_{I'}\mathbb{Z}$.

If that is not the case, then (identifying $I$ with $\mathbb{N}$) for any $n$, we find $v_n\in \prod_{\mathbb{N}} \mathbb{Z}$ with $v_n(i)=0$ for $i<n$ and $f(v_n)\neq 0$. But then we may consider sequences $a_i$ with $a_i\mid a_{i+1}$, and form $x_n = \sum_{i=0}^n a_i v_i$. Note this also makes sense for $n=\infty$. Now observe that $f(x_n) = f(x_\infty)$ modulo $a_{n+1}$, but if we inductively fix the elements $a_n$, we may arrange for the absolute value of the minimal representative of $f(x_n)$ modulo $a_{n+1}$ to diverge, leading to a contradiction. Simply pick $a_n$ large enough so that $f(x_n)$ is larger than $f(x_i)$ for all $i<n$, and then ensure that additionally $|a_{n+1}| > 2|f(x_n)|$ when we pick $a_{n+1}$.

Why does this help? If $M$ is a retract of $\prod_I \mathbb{Z}$, then $\operatorname{Hom}(M,\mathbb{Z})$ is a retract of $\bigoplus_I \mathbb{Z}$, hence free, and so $\operatorname{Hom}(\operatorname{Hom}(M,\mathbb{Z}),\mathbb{Z})$ is a product of copies of $\mathbb{Z}$. In general, this does not agree with $M$, however, in this situation it does! We always have a natural transformation $M\to \operatorname{Hom}(\operatorname{Hom}(M,\mathbb{Z}),\mathbb{Z})$, let us call an abelian group "reflexive" if this is an isomorphism. The above argument shows that $\prod_I \mathbb{Z}$ is reflexive, and since retracts of isomorphisms are retracts, it follows more generally that every retract of $\prod_I \mathbb{Z}$ is reflexive.

  • $\begingroup$ For the statement in your first paragraph, I think you need $|I|$ to be smaller than the first measurable cardinal (if such a thing exists). $\endgroup$ Jun 9 at 14:55
  • $\begingroup$ What do you mean by "dualize"? If you mean Potryagin Duality, I don't see how that helps. If your dualizing object is $\mathbb Z$, I see that you get that $Hom(G,\mathbb{Z})$ is a free abelian group. But how does that prove the original statement? $\endgroup$
    – Echo
    Jun 9 at 15:08
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    $\begingroup$ I just edited my answer: I addressed the question how the dualizing argument works, but unfortunately I also observed that the countable case does not formally imply the case of general $I$. So for now this is only an answer for countable $I$ (and maybe one can extend it somehow to the generality addressed by @JeremyRickard?) $\endgroup$ Jun 9 at 15:39
  • $\begingroup$ @AchimKrause (To complement Jeremy's comment) if you precisely need that the canonical map $\mathbf{Z}^{(I)}\to\mathrm{Hom}(\mathbf{Z}^I,\mathbf{Z})$ is an isomorphism, then you precisely need the assumption that $I$ has cardinal $<$ than the smallest measurable cardinal (if it exists) — equivalently that every ultrafilter on $I$, $\sigma$-complete (=stable under countable intersections) is principal. This is true for $I$ countable, but also $I$ continuum, power of the continuum, and so on. $\endgroup$
    – YCor
    Jun 11 at 7:10
  • $\begingroup$ If $\eta$ is a $\sigma$-complete ultrafilter on $I$, then $f\mapsto \lim_\eta f$ is a well-defined homomorphism $\mathbf{Z}^I\to\mathbf{Z}$. If $\eta$ is not principal, then this has "infinite support". $\endgroup$
    – YCor
    Jun 11 at 7:12

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