Can a non-compact group have a (non-trivial) compact subgroup?

Question

Yeah, the title says it all. It'll be more helpful if someone could provide with some canonical examples in case of an affirmative answer.

EDIT: Those suggesting the trivial subgroup - my mistake - I was after a non-trivial answer. updated the OP. Thanks!

Answer 1

The answer is yes.

Example. Let $G:=\mathrm{GL}(n,\mathbb{R})$, then $H:=O(n)$ is a compact subgroup of $G$ which is non-compact. Indeed, for all $k\geqslant 1$, $\textrm{diag}(k,1,\ldots,1)\in G$, whence $G$ is non-bounded and therefore non-compact. Besides, by definition $H:=\varphi^{-1}(\{I_n\})$ where: $$\varphi(A)={}^\intercal AA$$ is continuous as a quadratic form. Hence, $H$ is closed and its bounded by definition, whence compact.

Actually, all maximal compact subgroups of $G$ are conjugates of $H$. Another way to put it, any compact subgroup of $G$ is a conjugates of a subgroup of $H$.

Answer 2

The unit circle of the complex numbers (multiplicative group with $0$ removed) is a compact subgroup.

Similarly, $\{-1,+1\}$ is a compact subgroup of the nonzero real numbers.

Answer 3

Here is a whole family of examples: The matrix groups $\mathrm{SL}(n,\mathbb{C})$, $\mathrm{GL}(n,\mathbb{C})$, $SO(n,\mathbb{C})$, and $\mathrm{Sp}(n,\mathbb{C})$ are all real Lie groups. They are not compact (I think this is pretty easy to see). Each has a maximal compact subgroup though. In order, they are $\mathrm{U}(n)$, the unitary matrices, $\mathrm{SU}(n)$ the special unitary matrices, $\mathrm{SO}(n)$ the special unitary matrices with real entries, and $\mathrm{Sp}(n)$, which is $\mathrm{Sp}(n,\mathbb{C})\cap\mathrm{U}(2n)$.