A compact infinite topological group with only two closed subgroups It can be proved every compact infinite abelian topological group $(A,\tau )$, with $\tau$ nontrivial, has at least three distinct closed subgroups.
Is there any compact infinite non-abelian topological group $(G,\mathcal T)$, with $\mathcal T$ nontrivial, which has only two closed subgroups?
 A: I suppose that the group is Hasdorff: otherwise consider the factor group with respect to the closure of $\{ 1\}$. Closed subgroups of the quotient will correspond to closed subgroups of the initial group.
Let $G$ be a non-abelian Hausdorff topological group. Then for all $g \in G$
$$C_G(g) = \{ x \in G : xg=gx\}$$
is a closed subgroup of $G$.
In particular, since $G$ is not abelian, there exist $g,h \in G$ such that $gh \neq hg$. So, $C_G(g)$ and $C_G(h)$ are two distinct closed  non-trivial proper subgroups of $G$ (since $g \in C_G(g) \setminus C_G(h)$ and $h \in C_G(h) \setminus C_G(g)$).
So, it seems that $G$ has at least 4 closed subgroups.
A: Every infinite locally compact group $G$ has infinitely many closed subgroups.
(1) $G$ is not torsion. Hence, we can suppose that $G$ has a dense, infinite cyclic subgroup $C$. If $G$ is discrete then $G=C$ all subgroups of $C$ are closed, there are infinitely many. Otherwise $G$ is compact. Its Pontryagin dual is discrete infinite, and hence has infinitely many subgroups, which by duality provides as many closed subgroups in $G$.
(2) $G$ is torsion. Since $G$ is infinite and union of its cyclic subgroups, which are all finite, it has infinitely many finite cyclic subgroups, which are closed.

I'm not sure whether the result (that there are infinitely many closed subgroups) holds, say, for Polish groups. The above shows that it's enough to assume that $G$ is abelian with a dense infinite cyclic subgroup. In addition, by a dévissage argument, we can assume that the only closed subgroups of $G$ are $\{0\}$ and $G$.
Assuming only Hausdorff, there exists an infinite Hausdorff topological group $G$ with only closed subgroups $\{0\}$ and $G$. Namely, take any dense cyclic subgroup $\langle c\rangle$ of the circle group, with the induced topology. Being cyclic, its nontrivial subgroups all have the form $\langle c^n\rangle$ for some $n\ge 1$, but all these are dense.
