I asked myself this question recently, and I can't find the answer :
Let $G$ be a Polish group such that for every closed normal subgroup $H$ of $G$, $H$ is countable. Is $G$ necessary equal to $\mathbb{R}$ or $\mathbb{S_1}$ ? It is obviously the case if $G$ is a Lie group but I dont find any proof or counterexample in the setting of Polish groups.
No. There are Polish groups which are even algebraically simple. Whenever such a group is not locally compact, of course it cannot be $\mathbb{R}$ or $\mathbb{S}^{1}$.
An example to an algebraically simple non-locally compact Polish group is $\mathrm{Diff}_{o}^{\infty}(\mathbb{S}^{1})$, the identity component in the group of diffeomorphisms of the circle $\mathbb{S}^{1}$.
