Matrix Lie group counter-example: $e^X$ in the Lie group, but $X$ is not in the Lie algebra

Question

What's an example of a Lie group $G$ and matrix $X$ such that $e^X \in G$ but $x \notin \mathfrak{g}$, where $\mathfrak{g}$ is the associated Lie algebra?

This is the same as problem 2.10 in Bryan Hall's "Lie Groups, Lie Algebras, and Representations: An Elementary Introduction."

Answer 1

Take $H = \{ \pm I\}$ that has trivial Lie algebra because it is a finite group. The matrix

$$X = \left(\begin{array}{cc} 0 & -\pi \\ \pi & 0 \end{array}\right)$$

is such that $e^X = \left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right)$ but $X$ is not zero and hence is not in the Lie algebra.

Answer 2

Old question, but another example with nontrivial Lie algebra: Take $G = \mathrm{SL}(n,\mathbb C)$, with Lie algebra $\mathfrak{g} = \mathfrak{sl}(n,\mathbb C) = \left\{X \in M_n(\mathbb C) : \mathrm{tr}(X) = 0\right\}$. Then $X := 2\pi i I_n$ does not lie in $\mathfrak{sl}(n,\mathbb C)$, but $e^X = I_n \in \mathrm{SL}(n,\mathbb C)$.