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We know how to tell if a topological group is a Lie group: this was famously asked by Hilbert and answered gloriously by Gleason, Montgomery and Zippin in the 50s (a locally compact topological group can be turned into a Lie group iff there is one neighborhood of 1 which does not contain a non-trivial subgroup)

Can one tell when an abstract group is in some way a Lie group?

A natural follow up questions is, of course: if a group is Lie-sable (urgh), is it in a unique way? For comparison, a topological group which can made into a Lie group can be made in exactly one way.

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Generally it's not a Lie group in a unique way -- the groups $\mathbb R^n$ for $n \geq 1$ are all isomorphic as groups, but not as Lie groups with their standard Lie groups structures. –  Ryan Budney May 1 '11 at 19:58
Doesn't algori's answer to mathoverflow.net/questions/62385/… (and Kramer's paper I linked to in a comment there) at least partially answer your follow-up question? –  Theo Buehler May 1 '11 at 20:03
@Mariano, your question is very close to a standard question in group theory, being the question of if a group is linear i.e. if it is realizable as a subgroup of a matrix group. For this there's the Tits alternative. en.wikipedia.org/wiki/Tits_alternative –  Ryan Budney May 1 '11 at 20:18
Another condition: if $G$ is a (connected) Lie group, there is a number $k$ such that every torsion subgroup contains an abelian normal subgroup of index at most $k$ (This extends Jordan's bound for $GL$; see [Dong Hoon Lee: On Torsion Subgroups of Lie Groups. Proc. Amer. Math. Soc. Vol. 55, No. 2, 1976 ]) –  Mariano Suárez-Alvarez May 2 '11 at 3:14
In a Lie group $G$, every $g$ sufficiently close to the identity has the property that, for all positive integers $n$, there is an $h$ with $h^n=g$. The subgroup generated by all such $g$ must then contain the connected component of the identity, and so has at most countable index in $G$. For example, this shows that $\mathbb{Z}_p$ (forgetting the topology) cannot be given the structure of a Lie group. (This example is maybe not so interesting, as every torsion-free abelian Lie group is $\mathbb{R}^n$ for some $n$, but oh well) –  Julian Rosen May 2 '11 at 8:05

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