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Assuming the Hilbert-Smith conjecture, $G$ is a Lie group. For sufficiently large $n$ (I think $n=5$ will do) any $n$-connected Lie group is unipotent. But $G$ is not unipotent, because it contains $GL_m$. So no such $G$ can exist.


Here is a partial positive answer, going in the opposite direction of Terry┬┤s comment. Let $X$ be an infinite topological space. Suppose $X$ is first countable at $p \in X$ and for every open neighborhood $U$ of $p$ there is a smaller neighborhood $V \subseteq U$ such that $|U \setminus V|=|U|$. Then there is a group structure on $X$ with identity $p$ such ...


Having slept on it, I realise that the answer is No (even if we omit the condition on the inverse). The first example of infinite countable non-topologizable group due to Olshanskii (1980) is actually locally non-topologizable. Indeed, Olshanskii's group $G$ has exponent $p^2$ and the cyclic centre of order $p$. In addition, $g^p\ne1$ for any non-central ...

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