Suppose $G_1, G_2, \dots, G_n, \dots$ are groups (I use countable sequences, though the question is also applicable for uncountable collections of groups). Suppose G is the unrestricted external direct product of the $G_i$s. What is the derived subgroup (commutator subgroup) of G?

I get two things:

  1. The derived subgroup of G contains the restricted direct product of the derived subgroups of the $G_i$s, i.e., the set of elements with finitely many nonzero coordinates where each coordinate is in the derived subgroup of the corresponding group.
  2. The derived subgroup of G is contained in the unrestricted direct product of the derived subgroup of the $G_i$s, i.e., the set of elements where all coordinates are in the derived subgroups of their respective groups.

A more precise characterization of the derived subgroup seems to be: the set of those elements for which there exists some finite number c (depending on the element) such that every coordinate can be expressed as a product of at most c commutators in its coordinate group $G_i$.

Is this characterization correct? Also, under what conditions does the derived subgroup become either of the two extremes (the restricted direct product, or the unrestricted direct product)?

Also, the above suggests that there could be examples where all the $G_i$s are perfect groups [ADDED: A perfect group is a group that equals its own derived subgroup; however, it is not necessary that every element be a commutator] but their unrestricted direct product is not a perfect group. Could somebody come up with an example of this sort?

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    $\begingroup$ The way to get an unrestricted product of perfect groups, which is not perfect, is to choose $G_i$ such that every element in $G_i$ is the product of at most $i$ commutators, and there exists an element which cannot be written as $i-1$ commutators. Then the product over all $G_i$ will not be perfect. $\endgroup$ – Steve D Aug 12 '10 at 21:14
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    $\begingroup$ Actually, the key to the above is that the commutator width of $G_i$ be at least $i$. There are infinite simple groups with infinite commutator width (even f.p. examples), so choose one called $G$. Then the infinite product of $G$'s is not perfect. $\endgroup$ – Steve D Aug 12 '10 at 21:53

I believe your characterization is right.

If you take all the $G_i$ to be finite nonabelian simple groups, then Ore's conjecture implies that every element in each $G_i$ is a commutators. The conjecture was completed recently by Liebeck, O'Brien, Shalev, and Tiep, see http://www.math.auckland.ac.nz/~obrien/research/ore.pdf. So in this case your $G$ is actually perfect. (Actually this follows from weaker results then the full conjecture, which were known before).


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