Suppose $G$ is a locally Hausdorff topological group,does there must exits a non-trivial compact subgroup?
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José Carlos Santos
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math112358
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One can completely classify connected locally compact Hausdorff groups which have only trivial compact subgroups: These are simply-connected solvable Lie groups (of which ${\mathbb R}$ is the simplest nontrivial example) and their semidirect products with the universal cover of $SL(2,{\mathbb R})$. On the other extreme, if you consider discrete groups, then having no nontrivial compact subgroups is equivalent to being torsion-free. See also https://math.stackexchange.com/questions/475385/under-what-conditions-is-the-exponential-map-on-a-lie-algebra-injective/1592257#1592257 – Moishe Kohan Oct 03 '19 at 16:16
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To literally answer the question, the trivial group ${1}$ is already an example showing the answer is "no". It's always a good idea to check trivial cases before asking a question. – YCor Oct 04 '19 at 20:18
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No. Take $(\mathbb R,+)$, for instance. If $H$ is a subgroup of $(\mathbb R,+)$ and $H\neq\{0\}$, then $H$ contains some element $h\neq0$. But then $H$ will contain $2h$, $3h$ and so on and therefore $H$ will be unbounded. In particular, it will not be compact.
José Carlos Santos
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I have another quesion:can we classify the locally comapct topological group? – math112358 Oct 04 '19 at 10:36
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1That is another question. I suggest that you post it as such. – José Carlos Santos Oct 04 '19 at 11:01
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Say that a locally compact group is topologically torsion-free if it has no non-trivial compact subgroup. Elaborating on Moishe's comment, one sees that
A locally compact group $G$ is topologically torsion-free iff $G$ is Lie, the discrete quotient $G/G^\circ$ is torsion-free, and $G^\circ$ is contractible.
[And $G^\circ$ contractible means isomorphic to $S^k\ltimes R$ for some $k\ge 0$, some simply connected solvable Lie group $R$, and $S=\widetilde{\mathrm{SL}_2(\mathbf{R})}$].
To see this: any locally compact group $G$ has a connected-by-compact open subgroup $H$. If $G$ is topologically torsion-free, so is $H$. But every connected-by-compact locally compact group is compact-by-Lie (solution to Hilbert 5th problem). Hence $H$ is Lie, so $G$ is Lie too. From the connected Lie case, $G^\circ$ has the given form.
Now $G/G^\circ$ is torsion-free: indeed, otherwise, it has a nontrivial finite subgroup $L/G^\circ$. As a virtually connected Lie group, $L$ has a compact subgroup $K$ such that $KL^\circ=L$ (Mostow). Since $L$ is topologically torsion-free, $K=1$, and it follows that $L$ is connected, a contradiction. So $G/G^\circ$ is torsion-free.
Particular case (which can be checked directly using Pontryagin duality):
A locally compact abelian group is topologically torsion-free iff it is isomorphic to $\mathbf{R}^k\times \Lambda$ for some discrete torsion-free abelian group $\Lambda$ and some $k\ge 0$.
