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 $\vec p=(p\_k)\_{k\in\omega}$ of prime numbers, consider the Polish group $F\_{\vec p}=\prod\_{k\in\omega}F\_{p\_k}$.



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> **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 $\vec p=\vec q$? 
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The rigidity of the countable product of free groups
Yes, and even in the group category. More generally, I claim that if $\prod\_{i\in I}G\_i$ and $\prod\_{j\in J}H\_j$ 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 $f:I\to J$ 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 $H\subset\prod H\_i$, which has trivial intersection with the centralizer $K=\prod K\_i$ of $H$. Since by assumption $G=H\times K$, we deduce that $H=\prod H\_i$. Since $H$ is directly indecomposable, $H\_i$ is nontrivial for a single $i$. So $H\subset H\_i$, and $G\_i=H\_i\times K\_i$. Since $G\_i$ is directly indecomposable, we deduce that $H=H\_i=G\_i$.


