Are all matrix groups Lie groups?

Question

I am reading Lectures on Lie Groups and Lie Algebras, and I come across the statement on page 50 that says "All matrix groups are Lie groups". Does it mean all subgroups of general linear groups? If it does, how do I prove it?

Answer 1

The authors clearly goofed. There are even examples of matrix groups which are not isomorphic to Lie groups as abstract groups. (In the theory of topological groups one usually uses homeomorphic isomorphisms. When I say that an isomorphism of topological groups is abstract, I mean that it is a purely group-theoretic isomorphism, which is not required to be continuous.)

For instance, the additive group $\widehat{{\mathbb Z}}_p$ of p-adic integers is isomorphic to a subgroup $G$ of the additive group of real numbers ${\mathbb R}$. This isomorphism can be seen as follows. The algebraic closure $\overline{{\mathbb Q}_p}$ of the field of $p$-adic numbers is isomorphic (as a field) to ${\mathbb C}$. This gives an embedding (as an abstract group) of $\widehat{{\mathbb Z}}_p$ in the additive group of complex numbers. The latter is isomorphic (as an abstract group) to the additive group of real numbers.

One can show that the additive group $\widehat{{\mathbb Z}}_p$ is not isomorphic (as an abstract group) to any Lie group. (My definition of a Lie group includes the 2nd countable assumption.) The key is that $\widehat{{\mathbb Z}}_p$ is a profinite group, which implies that every nontrivial subgroup of $\widehat{{\mathbb Z}}_p$ has an epimorphism to a nontrivial finite (cyclic) group. At the same time, if $H$ is a connected abelian Lie group, $H$ does not admit (even discontinuous) epimorphisms to finite nontrivial groups. Since $\widehat{{\mathbb Z}}_p$ is uncountable, no subgroup of countable index in it is isomorphic to a Lie group. However, every Lie group contains a connected Lie subgroup of countable index. Thus, $\widehat{{\mathbb Z}}_p$ is not isomorphic to any Lie group (as an abstract group).