Does every infinite group admit a Hausdorff topology such that the multiplication and inverse are continuous at $1$ but $1$ is not an isolated point?
The question is inspired by and related to this one.
Surely, globally non-topologizable infinite groups do exist.
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 element of the group. So, if we have a topology on $G$ and the multiplication is continuous at $1$, then the function $f\colon x\mapsto x^p$ is also continuous at $1$. But $f$ takes only finitely many values ($p$ values). Thus, this continuous at $1$ function to a discrete (finite) set must be constant on a neighbourhood of $1$ and, hence, this neighbourhood must be finite (because the $f(x)=1$ only for central $x$).
Details of Olshanskii's construction can be found if his book.
