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Several days ago, Professor Martin Bridson gave a very nice talk in my department. Some questions concerning his talk came into my brain Since I am neither a model theorist nor a algebraist, I am not sure whether they fit here or not.

Question 1. Given a finite group $Q$, is there a sentence $\varphi$ in the group language so that for any group $G$, $G\models \varphi$ if and only if there is a surjective homomorphism from $G$ to $Q$?

Now let $G$ be a residually finite group and the corresponded inverse system $A=\{G/N\mid G/N \mbox{ is finite}\}$. Define $\hat{\Gamma}$ to be the inverse limit of $A$. Martin only focused on countable group and he mentioned that there is a quite important conjecture which says that for any two countable residually finite groups $G_0$ and $G_1$, $\hat{G_0}\cong\hat{G_1}\implies G_0\cong G_1$.Now I have the following question.

Question 2. Is it true that for arbitrary residually finite groups $G_0$ and $G_1$ (they do not need to be countable), $\hat{G_0}\cong\hat{G_1}\implies G_0\equiv G_1$, where $\equiv$ means elementary equivalence relation?

I am also interested the relationship between $G$ and $\hat{G}$. For example, $G\equiv_1 \hat{G}?$, where $\equiv_1$ means that they satisfy exactly same $\Sigma_1$-formulas?

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    $\begingroup$ You would at least want your group to be residually finite for the second question since the profinite completion of any infinite simple group is isomorphic to the profinite completion of the trivial group. $\endgroup$ Mar 25 '16 at 23:05
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    $\begingroup$ Do you mean homemorphism, or homomorphism? In the former case you'd need a topology, and you probably want to the take discrete one, in which case you just mean isomorphism, and in that case the answer is trivially yes. Just write the multiplication table of $Q$. In the latter case, you probably want it to be surjective, or at least nontrivial, otherwise every group has a trivial homomorphism into $Q$, so trivially $\exists x(x=x)$ works. $\endgroup$
    – Asaf Karagila
    Mar 25 '16 at 23:05
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    $\begingroup$ Semidirect products $\mathbf{Z}/11\mathbf{Z}\rtimes_2\mathbf{Z}$ and $\mathbf{Z}/11\mathbf{Z}\rtimes_8\mathbf{Z}$ are non-isomorphic with isomorphic profinite completion. I don't know if they are elementary equivalent. $\endgroup$
    – YCor
    Mar 25 '16 at 23:26
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    $\begingroup$ It's very likely false, even when $Q$ has order 2. We'd need one group $G$ in which every element is a product of squares, but for no $n$ every element is a product of $n$ squares. $\endgroup$
    – YCor
    Mar 26 '16 at 2:04
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    $\begingroup$ The examples I mentioned with $Z/11$ are indeed elementary equivalent EE. Indeed Oger (1982 ams.org/mathscinet-getitem?mr=689074, Théorème 5.5) proved that two finite-by-abelian f.g. groups $G,H$ (I mean, finite kernel, abelian quotient) are EE iff their profinite completions are isomorphic as topological groups (with other characterizations, e.g., that is equivalent to: $G\times\mathbf{Z}$ and $H\times\mathbf{Z}$ are isomorphic groups). (Note that abstract homomorphisms between top. f.g. profinite groups are continuous, by Nikolov-Segal, which is easy in this special setting). $\endgroup$
    – YCor
    Mar 26 '16 at 10:27

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