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

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."

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Take $G$ to be trivial... –  Qiaochu Yuan Oct 5 '12 at 1:14

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.
(For what it's worth, Maple says that $e^X = -I$, but you can just change $H$ to accommodate it. In order to see why your answer must be wrong, note that the nontrivial element of $H$ has determinant $-1$, so is in a different connected componenet of $GL_2$ than $I$. In particular, it cannot be $e^X$ for any $X$). –  Jason DeVito Oct 5 '12 at 2:16