Why aren't Lie Algebra generators invertible if we apply constarints from the Lie Group? We often find the generators of a Lie Algebra by applying the constraints of the Lie Group up to linear order in the generator $T$. For example, for $SO(7) = \{S \in GL(R^4) | S^TS=1, \det(S)=1 \}$ we have:
Constraint 1 - $S^TS=1$:
$$(1+T)^{T} (1+T) = 1 \implies T^T = -T$$ i.e. $T$ must be anti-symmetric
Constraint 2 - $\det(S)=1$:
$$det(1+T) = 1 + tr(T) +O(T^2) = 1 \implies tr(T)=0$$ i.e. $T$ must be traceless (already satisfied)
My question is why do we not also include the $S \in GL(R^4)$ constraint to therefore make $1+T \in GL(R^4)$ and therefore $T \in GL(R^4)$ instead of $T \in End(R^4)$
I understand that Lie Algebra elements need not be invertible while Lie Group elements must, but if we are defining the Lie Algebra elements to be isomorphic to the generators of the Lie Group then shouldn't we have to apply this constraint too?
 A: 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 $-T$.
