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It is well-known that the direct product of any family of abelian groups is an elementary extension of the direct sum of the family (see e.g. Lemma A.1.6 in the book `Model Theory' by W. Hodges, where this is proven for modules). Can this result be generalized to some other classes of groups? More concrete questions:

(1) Is there a family of groups such that its direct product is not elementarily equivalent to its direct sum? Are there a group $G$ (a finite group $G$) and a cardinal $\kappa$ such that $G^{\kappa}$ is not elementarily equivalent to $G^{(\kappa)}$?

(2) Is there a family of groups such that its direct product is elementarily equivalent to its direct sum but is not an elementary extension of it? Are there a group $G$ (a finite group $G$) and a cardinal $\kappa$ such that $G^{\kappa}$ is elementarily equivalent to $G^{(\kappa)}$ but is not an elementary extension of it?

(3) Are there classes of groups, becides the classes of abelian groups, in which direct products are elementary extensions of direct sums?

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  • $\begingroup$ What is the direct sum of nonabelian groups? $\endgroup$
    – Tim Campion
    Commented May 17, 2015 at 0:05
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    $\begingroup$ @Tim: it's the subgroup of the direct product where all but finitely many terms vanish. (Its universal property is that it is the "commutative coproduct": it's universal with respect to having maps in from the groups whose images commute.) $\endgroup$
    – Qiaochu Yuan
    Commented May 17, 2015 at 0:14
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    $\begingroup$ Confusingly there is an alternative convention in which "direct product" refers to the finite support version and "cartesian product" to the unrestricted version. $\endgroup$
    – Derek Holt
    Commented May 17, 2015 at 10:51

1 Answer 1

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Consider the sentence $P:\forall x\exists y:y^3=1\neq y,yx=xy$. Let $(F_i)$ be an infinite family of groups, each of which possesses an element of order 3. Then $\bigoplus F_i$ satisfies $P$.

Now assume that each $F_i$ is non-abelian of order 6: in $F_i$, elements of order 2 never commute to elements of order 3. Thus in the direct product, an element all of whose components have order 3 commutes with no element of order 2. This answers (1).

Also when each $F_i$ is non-abelian of order 8, the direct sum satisfies ``each element has a non-abelian centralizer" (which can be made 1st order), but not the direct product.

More generally, if there exists $n_0$ such that in each $F_i$, there exists a family of $n_0$ elements whose centralizer in $F_i$ is abelian (this obviously holds if $F_i$ has bounded cardinal, but also more generally if $F_i$ has finite generating rank (e.g. if $F_i$ is simple since then it is generated by 2 elements). Then the direct product $\prod F_i$ contains a finite family whose centralizer is abelian; while the direct sum $\bigoplus F_i$ has such a finite family only if all but finitely many $F_i$'s are abelian. In this very general case again, the direct sum and the direct product are not elementary equivalent, and when $F$ is finite non-abelian no infinite $F^{(\alpha)}$ can be elementary equivalent to any $F^\kappa$.

Since I don't know a single case for which infinitely of the $F_i$'s are non-abelian and I know the direct sum to be elementary equivalent to the direct product, I have no idea about (2).

Edit: It took me some time to struggle until I solved the following: in a group $F$, let $m(F)$ be the smallest cardinal of a subset of $F$ whose centralizer is abelian. Find finite groups with $m(F)$ arbitrary large (even finding $F$ with $m(F)\ge 3$ did not seem immediate). Note that $m(A\times B)=\max(m(A),m(B))$, so direct products do not help; $m(F)$ is bounded above by the minimal number of generators, and is 0 for abelian groups. Actually, if we fix a prime $p$ and consider the product of a large family of $n$ non-abelian groups of order $p^3$ and glue their center, so that the resulting group $G$ has order $p^{2n+1}$ and has both its center and its derived subgroup of order $p$, then the centralizer of any noncentral element has index $p$, and hence the centralizer of any family of $k$ elements has index $\le p^k$; on the other hand this group has no abelian subgroup of order $>p^{n+1}$, so $m(G)\ge n+1$ (actually it's an equality). This provides families of $(F_i)$ for which the previous argument does not work.

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  • $\begingroup$ Your argument also shows that G^k and G^(k) are not elementarily equivalent, for finitely generated non-abelian groups G and infinite k: if G is n-generated then the first-order statement 'there are n elements with abelian centralizer' is true in G^k and false in G^(k). This gives information on Question 3: if a class of groups has the property 'G^k is elementarily equivalent to G^(k) for all G and k' then all finitely generated groups in the class are abelian, and hence a class of groups which is closed under subgroups has this property if and only if all groups in the class are abelian. $\endgroup$
    – owb
    Commented May 17, 2015 at 16:14
  • $\begingroup$ So, it seems the following weak version of Question 2 remains open: Is there a non-abelian group G such that G^k and G^(k) are elementarily equivalent, for some infinite cardinal k? As Yves' argument shows, such G cannot be finitely generated. $\endgroup$
    – owb
    Commented May 17, 2015 at 22:50
  • $\begingroup$ I also don't know if there exists a non-abelian group $G$, of continuum cardinal, such that $G^{\aleph_0}$ and $G^{(\aleph_0)}$ are isomorphic; note that they both have continuum cardinal. (In the abelian case there are both examples for which they are isomorphic and non-isomorphic.) $\endgroup$
    – YCor
    Commented May 17, 2015 at 23:29

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