
> 
> 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?
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The question is inspired by and related to [this one](https://mathoverflow.net/q/173391/24165).


Surely, globally non-topologizable infinite groups [do](https://mathoverflow.net/a/165787/24165) [exist](https://mathoverflow.net/q/114688/24165).



Are infinite groups "locally topologizable"?
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](http://books.google.ru/books?id=jwNkqFQn-GcC&lpg=PP1&dq=olshanskii%20geometry%20of%20defining&pg=PP1#v=onepage&q&f=false). 


