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YCor
YCor
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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 $\vec p=(p_k)_{k\in\omega}$ of prime numbers, consider the Polish group $F_{\vec p}=\prod_{n\in\omega}F_{p_k}$$F_{\vec p}=\prod_{k\in\omega}F_{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 $\vec p=\vec q$?

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 $F_{\vec p}=\prod_{n\in\omega}F_{p_k}$.

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 $\vec p=\vec q$?

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 $F_{\vec p}=\prod_{k\in\omega}F_{p_k}$.

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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Taras Banakh
Taras Banakh
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The rigidity of the countable product of free groups

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 $F_{\vec p}=\prod_{n\in\omega}F_{p_k}$.

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 $\vec p=\vec q$?