Is there a topological group with the small index property that does not have automatic continuity?

Question

Here are the exact definitions of the terms:

Let $G$ be a topological group. Then $G$ has the small index property if every subgroup of countable (including finite) index is open in $G$. Furthermore, $G$ has automatic continuity if every group homomorphism from $G$ to any separable topological group is continuous.

It is well-known that $G$ having the small index property is equivalent to the fact that every homomorphism from $G$ to the symmetric group $S_{\infty}$ on a countably infinite set $X$ is continuous, where $S_{\infty}$ is given the usual pointwise topology with sub-basis $\{f\in S_{\infty}: f(x)=y\}$ over all points $x,y \in X$.

Hence automatic continuity of $G$ implies the small index property. Is an example known that shows the converse is false?

Answer 1

Here is a counterexample. Consider $\mathbb{R}$ with addition. We define a topology on this group by giving the cosets of all countable index subgroups as a sub-basis.

This subbasis is actually a basis as the intersection of finitely many subgroups of countable index is another subgroup of countable index. Moreover as $\mathbb{R}$ is abelian we can see this is a group topology, as if G is a countable index subgroup then $(G+x) + (G+y) \subseteq (G+x+y)$ and $-(G+x)= (G-x)$ for all $x,y \in \mathbb{R}$. This topolgical group has the small index property by the definition of the topology.

To see it doesn't have automatic continuity, consider the idenity map from the reals with this topology the the reals with the standard topology. This is a homomorphism (even an isomorphism), but the preimage of $(-1, 1)$ is itself. This is not open in our new topology as all open sets in our new topology contain a translation of a non-trivial group, and are thus unbounded.