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What can be said about the class of groups which can be represented as a homomorphic image of an (infinite) Cartesian product (=unrestricted direct product) of finite groups? What would be simple groups in this class? Which "classical" groups, like various matrix groups, belong to this class?

The question can be (trivially) reformulated as a description of the smallest class of groups containing all finite groups and closed under taking Cartesian products and homomorphic images.


  1. This class includes ultraproducts of finite groups, or, what is the same, groups having the same first-order theory as a set of finite groups (called pseudo-finite in the literature).
  2. A countable Cartesian product of finite groups is a profinite group. I don't know how beneficial it would be to consider the question in the profinite context.
  3. A finitely-generated group from this class is necessarily finite (Nikolov and Segal, see arXiv:1108.5130, Theorem 32).
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@Pasha: Group having the same first order theory as a finite group $G$ is isomorphic to $G$ (because size $n$ is a first order sentence). You probably meant "having the same first order theory as a set of finite groups". – Mark Sapir Apr 3 '12 at 7:31
Also you probably meant "Cartesian product". – Mark Sapir Apr 3 '12 at 7:33
@Mark: "same first order theory": thanks, of course, corrected it. "Cartesian product": it seems that I am confused about the terminology. Aren't "Cartesian product" and "direct product" the same, as opposed to "direct sum"? – Pasha Zusmanovich Apr 3 '12 at 7:44
@Pasha: If you consider non-Abelian groups, there is no "direct sum". Direct product consists of functions with finite supports, Cartesian product consists of all functions. – Mark Sapir Apr 3 '12 at 11:58
Cartesian product is a product of sets, and in universal algebra, the product structure whose domain is the Cartesian product is called direct product. Wikipedia… suggests that this agrees with the usage in group theory. – Emil Jeřábek Apr 3 '12 at 13:38

@Pasha, you might like to look at if my memory serves me right, you might find some answers to your questions.

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Thanks! (amended the question) – Pasha Zusmanovich Apr 3 '12 at 13:27

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