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For a natural number $n$ let $F_n$ be the free group with $n$ generators.
The group $F_n$ is endowed with the discrete topology.

Given an increasing sequence $vec p=(p_k)_kinomega$ of prime numbers, consider the Polish group $F_vec p=prod_kinomegaF_p_k$.

Problem. Let $vec p,vec q$ be two increasing sequences of prime numbers such that the Polish groups $F_vec p$ and $F_vec q$ are topologically isomorphic. Is $vec p=vec q$?

Yes, and even in the group category. More generally, I claim that if $prod_iin IG_i$ and $prod_jin JH_j$ are isomorphic groups, for two sets $I,J$ and two families $(G_i)$, $(H_j)$ of groups that are center-free and directly indecomposable, then there is a bijection $f:Ito J$ such that $G_i$ and $H_f(i)$ are isomorphic for all $i$.

(Recall that a group is directly indecomposable if it's nontrivial and not direct product of two nontrivial subgroups.)

To show this, it is enough to recognize the subgroups $G_i$ in the product $G=prod_i G_i$, purely relying on the structure of the group $G$.

Indeed, $G_i$ is directly indecomposable, and $G$ is direct product of $G_i$ and its centralizer (using that $G_i$ has trivial center).
I claim that this characterizes the subgroups $G_i$.

Claim if $H$ is a directly indecomposable subgroup of $G$ such that $G$ is direct product of $H$ and its centralizer, then $H=G_i$ for some unique $i$.

Trivial lemma: Let $H$ be a subgroup of $G=prod G_i$ ($G_i$ arbitrary groups), and $H_i$ its projection on $G_i$. Let $K_i$ be the centralizer of $H_i$. Then the centralizer of $H$ is $prod_i K_i$. $Box$

Now to prove the claim, let $H$ be a subgroup with these properties. Then $Hsubsetprod H_i$, which has trivial intersection with the centralizer $K=prod K_i$ of $H$. Since by assumption $G=Htimes K$, we deduce that $H=prod H_i$. Since $H$ is directly indecomposable, $H_i$ is nontrivial for a single $i$. So $Hsubset H_i$, and $G_i=H_itimes K_i$. Since $G_i$ is directly indecomposable, we deduce that $H=H_i=G_i$.