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.

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  • $\begingroup$ What about $G=\{e\}$? $\endgroup$ – Kavi Rama Murthy Jun 26 '19 at 10:28
  • $\begingroup$ It seems that the word “space” should be replaced by “group”. $\endgroup$ – Gabe Conant Jun 27 '19 at 1:49
  • $\begingroup$ Thanks for correction $\endgroup$ – user169239 Jun 27 '19 at 2:00
  • $\begingroup$ Are you aware of the classification theorem for locally compact groups (Gleason-Yamabe)? If not, read here: terrytao.wordpress.com/tag/locally-compact-groups. Once you understand it, you will answer your own question. Ping me if you cannot do this after reading Tao's post. $\endgroup$ – Moishe Kohan Jun 28 '19 at 15:01
  • $\begingroup$ I know this result sir but I don't know how to use. $\endgroup$ – user169239 Jun 29 '19 at 7:07

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.

  1. 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.

  2. 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.

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  • $\begingroup$ In the second case, is there an easy way to see that $K$ must be compact? $\endgroup$ – Eric Wofsey Jun 28 '19 at 20:22
  • $\begingroup$ Theorem 19(Page 71) in Sidney A. Morris book namely "Pontryagin duality and the structure of locally compact abelian groups " it follows that if G is locally compact group then closure of cyclic subgroup is isomorphic to Z or compact group. $\endgroup$ – user169239 Jun 29 '19 at 6:58
  • $\begingroup$ Thank you sir for your answer (YCor). $\endgroup$ – user169239 Jun 29 '19 at 7:10

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