the existence of a compact subgroup

Question

Suppose $G$ is a locally Hausdorff topological group,does there must exits a non-trivial compact subgroup?

Answer 1

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.

Answer 2

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 $k\ge 0$.