Do every infinite locally compact Hausdorff group has infinitely many closed subgroup? Do every infinite locally compact Hausdorff group has infinitely many closed subgroup?
I know that every locally compact compactly generated abelian group is topologically isomorphic to $\Bbb R^n\times \Bbb Z^m\times$ compact abelian group.
If $G$ is an infinite real Lie group than $G$ has infinitely many closed groups but in general, I don't know.
 A: Yes (no classification needed).
(a) first case: $G$ has a closed subgroup isomorphic to $\mathbf{Z}$: then OK (since $\mathbf{Z}$ has infinitely many subgroups).
(b) second case: $G$ has an element $g$ of infinite order (but no closed subgroup isomorphic to $\mathbf{Z}$). Then the closure $K$ of $\langle g\rangle$ is an infinite compact subgroup. Its Pontryagin dual is an infinite discrete abelian group, so has infinitely many subgroups by the discrete case, and again by Pontryagin duality it follows that $G$ has $K$ has infinitely many closed subgroups.
(c) if neither (a) nor (b) applies, then every element has finite order, so $G$ is covered by finite subgroups. Hence $G$ has infinitely many finite (hence closed) subgroups.

Let me provide a second way, a bit more self-contained.

Proposition: if a locally compact group $G$ has no infinite increasing or decreasing chain of closed subgroups, then it's discrete and finite.

Indeed, if we are not in the setting of (c) above, then some cyclic subgroup has an infinite closure and hence we can suppose that $G$ is abelian. Then using the assumption, there is a finite saturated chain of closed normal subgroups. Hence it is enough to prove the following.

Proposition: let $G$ be a locally compact group whose only closed subgroups are $\{0\}\neq G$. Then $G$ is discrete, cyclic of prime order.

Indeed, $G$ is clearly abelian. Also it is either connected or totally disconnected (since the connected component of $0$ is a closed subgroup). We conclude separately.


*

*if $G$ is totally disconnected, Dantzig's theorem says that $0$ has a system of neighborhoods consisting of open subgroups. Hence by the assumption, $G$ is discrete and this case is standard. Dantzig' theorem is very easy modulo proving that every Hausdorff locally compact space has at lease one nonempty compact clopen subset.

*if $G$ is connected, use Pontryagin duality only in the way that $G\neq 0$ implies the existence of a nontrivial homomorphism $f$ to the circle group $\mathbf{R}/\mathbf{Z}$. By connectedness, $f$ is surjective. So $f^{-1}((\frac12\mathbf{Z})/\mathbf{Z})$ is a nonzero proper subgroup, contradiction.
