If $G$ is a Lie Group (with identity element of $e$), then my definition of the Lie Algebra $\mathfrak{g}$ of $G$ is the tangent space of $G$ at $e$, so that $\mathfrak{g} = T_{e}G$.
The Lie Algebra of $both$ $\mathrm{SO}(2)$ and $\mathrm{O}(2)$ is given by:
$T_{e}\mathrm{O}(2) = T_{e}\mathrm{SO}(2) = \{ \ A \in M_{2 \times 2}(\mathbb{R})\ |\ A^{T} = - A \ \}$
My question is, WHY are they the same? My thinking is that since:
$\mathrm{O}(2) = \mathrm{SO}(2) \cup \{ \ A \in M_{2 \times 2}(\mathbb{R})\ |\ A^{T}A = I\ \mathrm{and}\ \det(A) = -1 \ \}$
Which means that $\mathrm{O}(2)$ is disconnected. Since one of its constituent disconnected sets is $\mathrm{SO}(2)$, then it must have the same Lie Algebra as this set? Is this kind of thinking correct? I am not sure if I have articulated myself properly.