# Is every countable discrete group a subgroup of a non discrete Lie group?

1)Let $$G$$ be a countable discrete group. Can $$G$$ be embbeded in a locally connected Lie group?

2)let $$G$$ be a countable discrete group with a prescribed generating set and corresponding word metric. Can $$G$$ be embbeded isometrically in a locally connected Lie group with its left invariant metric?

Remark: We emphasis on the word "Locally connected" because of the example $$G\subset \mathbb{R}\setminus\{0 \}$$ with $$G=\{\pm e^n\mid n\in \mathbb{Z} \}$$ with multiplication.

• I assume you want more conditions, or else you can just take $G \times \mathbb R$? – Kevin Casto Aug 13 '20 at 11:26
• In that case there is a huge literature on this subject. See e.g. this question: mathoverflow.net/questions/110208/… I suppose technically you need to worry about nonlinear Lie groups like $\widetilde{SL_2}(\mathbb R)$ but I suspect the properties listed in the first answer to that question also apply to discrete subgroups of this. – Kevin Casto Aug 13 '20 at 12:07
• Every Lie group is locally connected, so the answer to your question is obviously yes since every countable discrete group is a Lie group. On the other hand, many countable discrete groups do not embed (even algebraically) in connected Lie groups. – Moishe Kohan Aug 13 '20 at 13:31
• If you mean nondiscrete, then the firs comment by Kevin answers the question. As for examples, take any centerless nonresidually finite group such as BS(2,3), Baumslag Solitar group. Or take any group with undecidable word problem if you want a more exotic example. – Moishe Kohan Aug 13 '20 at 13:49
• $BS(2,3)=<a,b| ab^2a^{-1}=b^3>$ does not embed in a connected Lie group. – Moishe Kohan Aug 13 '20 at 14:02

Malcev proved that every finitely-generated matrix group $$\Gamma$$ (over any field) is residually finite, i.e. the intersection of all finite-index subgroups of $$\Gamma$$ is $$\{1\}$$. Baumslag-Solitar groups, such as $$BS(2,3)= \langle a, b | ab^2 a^{-1} =b^3\rangle$$, are among simplest examples of finitely generated groups which are not residually finite. A connected Lie group $$G$$ need not be linear (the universal covering group of $$SL(2, {\mathbb R})$$ is a standard example). However, the kernel of the adjoint representation $$Ad_G$$ of a connected Lie group $$G$$ is always contained in the center of $$G$$. Thus, if $$\Gamma< G$$ is a centerless subgroup, then the restriction of the adjoint representation $$Ad_G$$ to $$\Gamma$$ is faithful and, hence, $$\Gamma$$ is isomorphic to a matrix group. It is not hard to see that $$BS(2,3)$$ has trivial center. Thus, this group is not isomorphic to a subgroup of any connected Lie group. The same proof shows that $$BS(2,3)$$ is not isomorphic to a subgroup of a Lie group with finitely many components.
Remark. The standard definition of "locally connected" in topology is that every point should have a neighborhood basis consisting of connected subsets. Hence, each manifold (in particular, each Lie group) is, by definition, locally connected. Given your example, it seems that what you really had in mind is that a Lie group $$G$$ should have Alexandroff compactification $$G\cup \{\infty\}$$, such that $$\infty$$ admits a neighbourhood basis $$U_i$$ satisfying the condition that $$U_i\cap G$$ is connected. It is easy to see that this requirement is equivalent to the condition that $$G$$ is connected and 1-ended (equivalently, is neither compact nor a product of compact group with $${\mathbb R}$$). I am not sure what to call this property, let's name it ($$*$$). Then every discrete countable group $$\Gamma$$ embeds in a Lie group with property ($$*$$), e.g., $$G=\Gamma \times {\mathbb R}^2$$.
I would have posted this as a comment, but I have not yet unlocked that privilege. For a simpler example than Moishe Kohan's, the group $$\bigoplus_{\mathbb{N}} \mathbb{Z}/2\mathbb{Z}$$ does not embed in any connected Lie group, since a torsion subgroup of a connected Lie group $$G$$ is contained in a maximal compact subgroup of $$G$$ (this is a result of D. H. Lee), and it is easy to see that $$\bigoplus_{\mathbb{N}} \mathbb{Z}/2\mathbb{Z}$$ does not embed in $$\mathrm{U}(n)$$ for any $$n$$. Also, it follows from the Tits alternative for linear groups that any group that is not virtually solvable and does not contain a nonabelian free subgroup does not embed in a connected Lie group (or a Lie group with finitely many connected components).