# Intersection of all open subgroups vs. the intersection of all open normal subgroups

I am interested to know examples of topological groups $$G$$ for which the intersection $$\bigcap\{H\leq G\mid H\text{ open}\}$$ of all open subgroups of $$G$$ is the trivial subgroup but for which the intersection $$\bigcap\{N\trianglelefteq G\mid N\text{ open}\}$$ of all open normal subgroups is not the trivial subgroup.

Clearly (1) must be totally disconnected (2) cannot inject into a pro-discrete group by a continuous homomorphism and (3) it can't contain a topological subgroup isomorphic to $$\mathbb{Q}$$. I imagine that topological groups fitting this description exist and perhaps some are even important in some area I am not familiar with.

Does such a topological group exist? If so, is there an abundance of "standard" examples?

• It's actually a very common behavior! For instance any nondiscrete topologically simple group with a proper open subgroup works. Nontrivial semisimple $p$-adic groups work too, e.g., $SL_2(\mathbf{Q}_p)$.
– YCor
Aug 6 at 7:29

$$S_\infty$$, the group of all permutations of $$\mathbb{N}$$, has a neighborhood base of the identity of open subgroups. (In fact a Polish group with that property is isomorphic to a closed subgroup of $$S_\infty$$).

But without thinking about exactly which ones are open, $$S_\infty$$ has a very limited supply of normal subgroups outlined here: https://math.stackexchange.com/a/166472/29633

• $S_\infty$ is topologically simple: all nontrivial normal subgroups are dense.
– YCor
Aug 6 at 7:30

There are extreme examples of such behaviour, namely nonarchimedean (meaning that they have a basis at the identity consisting of open subgroups) Polish groups that are simple as abstract groups (meaning that they have no normal subgroups at all rather than just no closed ones).

First of all nonarchimedean Polish groups are plenty: groups of the form $$\mathrm{Aut}(M)$$ for a countable homogeneous structure $$M$$ are Polish, being isomorphic to a closed subgroup of $$S_\infty$$ and nonarchimedean, having a basis at the identity consisting of automorphisms pointwise stabilizing finite sets (indeed as the other answer mentions this is an iff).

Many such groups are known to be simple, the first example historically is probably the automorphism group of the random graph, shown to be simple by Truss, but you can look at the 2011 paper by Macpherson-Tent for many examples, historical remarks and general results on simplicity of such groups.

• The first example historically is certainly $S_\infty$ itself (Aut of "discrete" structure). Its topology was defined by Onofri in the late 1920s, and he classified normal subgroups too.
– YCor
Aug 6 at 7:34
• @YCor I mean the first example of a nonarchimedean Polish group that is simple as an abstract group rather than just topologically simple Aug 6 at 7:37
• OK sure, but that's quite independent of the OP's question. Anyway, simplicity of Homeo(Cantor) was proved by Anderson in 1958.
– YCor
Aug 6 at 7:43