Is $SO(n)$ actuallly the same as $O(n)$?

Question

$SO(n)$ is defined to be a subgroup of $O(n)$ whose determinant is equal to 1. In fact, the orthogonality of the elements of $O(n)$ demands that all of its members to have determinant of either $1$ or $-1$. Denote $so(n)$ and $o(n)$ to be the Lie algebras of $SO(n)$ and $O(n)$ respectively, then it can be shown $so(n)=o(n)$. This is due to the identity $$ \textrm{det }e^X = e^{\textrm{Tr }X} $$ In the current problem, $X$ is a member of $so(n)$ (or of $o(n)$). Now since all members of $o(n)$ are anti-symmetric matrices which have all its diagonal elements zero, their traces are all zero: $\textrm{Tr }X=0$. This implies that $\textrm{det }e^X = 1$, so all members of $O(n)$ have determinant $1$, none of them have determinant $-1$. This further means that $SO(n)=O(n)$. But I feel like I go wrong somewhere, please someone makes it clear.

Answer 1

The image of the exponential map from the Lie algebra of a Lie group to the group is contained in the connected component of the identity element. In the case of $O(n)$, which is not connected, this means that not all elements of $O(n)$ are the exponential of something in the Lie algebra.