Multiplative subgroup of lie algebra

Let’s start with subalgebra A⊂GL(N,R)A \subset GL(N, R) with commutator as Lie bracket. It has subset of invertable elements which form group GG.

I want to proof (if it’s true of course)

GG is Lie group
Lie algebra of GG is AA

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I don’t understand this question. Lie algebras don’t have invertible elements, as x^2 = 0 for all elements x. However, if you mean in the underlying associative algebra, then every element is invertible.
– David Towers
13 hours ago

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