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Recall that Goursat's Lemma has the following useful consequence. Let $G_1, G_2$ be finite groups with no common simple non-abelian quotients, and suppose $\gcd(|G_1^{\operatorname{ab}}|, |G_2^{\operatorname{ab}}|) = 1$, where superscript $\operatorname{ab}$ denotes abelianization. If $H \subset G_1 \times G_2$ is a subgroup with the property that the natural projections $H \to G_1$ and $H \to G_2$ are surjective, then $H = G_1 \times G_2$. A statement/proof of this version of Goursat's Lemma can be found in Lemma A.4 of Zywina's article ``Elliptic Curves with Maximal Galois Action on their Torsion Points'' (see http://www.math.cornell.edu/~zywina/papers/MaximalGalois.pdf).

I would like to obtain a similar version Goursat's Lemma in the following more general situation. Let $I$ be an at-most-countable index set, and let $\{G_i\}_{i \in I}$ be a collection of topological groups. Let the product group $G = \prod_{i \in I} G_i$ be equipped with the usual product topology (a base of opens is given by sets of the form $U_{i_1} \times \cdots \times U_{i_n} \times \prod_{i \neq i_1, \dots, i_n} G_i$, where $U_{i_j} \subset G_{i_j}$ is open for each $1 \leq j \leq n$). Perhaps a statement of Goursat's Lemma in this situation would be something like the following:

Suppose no two of the $G_i$'s have any common simple non-abelian quotients, and suppose further that no two of the $G_i$'s have any common abelian quotients (this is the analogue of saying that the abelianizations have coprime order). If $H \subset G$ is a closed subgroup with the property that the natural projections $H \to G_i$ are surjective for every $i \in I$, then $H = G$.

Is the above blocked statement true, or are further assumptions required to make it true? For example, would the statement hold if the groups $G_i$ were profinite?

Here is what I have found so far:

  • Simply inducting on the size of $I$ with Goursat's Lemma for finite products isn't going to get the desired result.
  • Even if I consider the projection $H'$ of $H$ onto a product of $G_{i_1}, \dots, G_{i_n}$ and argue that that $H' = \prod_{j = 1}^n G_{i_j}$ with Goursat's Lemma for finite products, I still do not immediately know that $H \supset H'$.
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    $\begingroup$ Just for the record: what you refer to as Goursat's Lemma is just a very special case. The orginal Lemma of Goursat gives a description of arbitrary subgroups of the product of two arbitrary groups. In fact, Goursat was mainly interested in the case of infinite groups. Reference: E. Goursat, Sur les substitutions orthogonales et les divisions régulières de l’espace, Ann. Sci. École Norm. Sup. (3) 6 (1889) 9–102 (see p. 47-48) $\endgroup$ – Friedrich Knop Jul 2 '16 at 6:56
  • $\begingroup$ In the finite group case, your conditions make sure that no two factors have isomorphic nontrivial quotients. This is no longer the case for arbitrary (discrete) groups. So do you impose any restrictions on your groups? $\endgroup$ – Friedrich Knop Jul 2 '16 at 15:38

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