Why aren't Lie Algebra generators invertible if we apply constarints from the Lie Group?

Solution 1:

The crucial flawed step is

$1+T \in GL(R^4)$ and therefore $T \in GL(R^4)$

There is no "therefore" here. $1+T$ being (multiplicatively) invertible does not imply anything about (multiplicative) invertibility of $T$.

As example, look at the Lie group of upper triangular unipotent matrices with $1$'s on the diagonal. Its Lie algebra is the set of strictly upper triangular matrices, i.e. now with $0$'s on the diagonal. All elements of the Lie group are invertible, while all of the Lie algebra are not. But note that "group = 1 + algebra", the algebra is a space "so small that adding it to the group identity stays within the invertible matrices". So in a way, the constraint you want to impose is empty: Everything close to the identity is invertible, regardless of whether its difference from the identity is invertible or not.

Actually, as user Nate points out in a comment, when you go from the group to the Lie algebra you are turning multiplication into addition: Again, the idea is that the multiplicative inverse of $1+T$ is $1-T$ (up to linear order). So, as Nate aptly say, elements in $G$ having multiplicative inverses corresponds to the Lie algebra being closed under $(-1) \cdot$, i.e. with $T$, the Lie algebra also contains its (additive!) inverse $-T$.