Proof that matrices of the form $e^X$ are a group under matrix multiplication I have searched but could not find the information.
It is well known that
$$
\exp \mathfrak{gl} \to GL^+
$$
is only surjective if $\mathfrak{gl}$ are the complex-valued matrices. If they are real valued, the map is not surjective.
But then from the Baker–Campbell–Hausdorff formula1, we know that there exists many matrices such that $e^Xe^Y=e^Z$.
Can we say that exponentials of matrices are a group under multiplication? Because $\forall X \forall Y: e^Xe^Y=e^Z$, as per the formula above?
 A: Following the comment of @Moishe Kohan:
Given a $2\times2$ matrix $M$, let $t=\text{tr}(M)$ be the trace of $M$, let $F=\text{tf}(M)$ be the trace-free part of $M$, so $M=\tfrac t2I+F$, and let $d=\det(F)$ be the determinant of $F$. A straightforward calculation (or Cayley-Hamilton) gives $F^2=-d\,I$, so the exponential is
$$\exp(M)=\exp(\tfrac t2)\exp(F)=\exp(\tfrac t2)\cdot\begin{cases}\cos(\sqrt d)\,I+\frac{1}{\sqrt d}\sin(\sqrt d)\,F,\quad d>0\\I+F,\quad d=0\\\cosh(\sqrt{-d})\,I+\frac{1}{\sqrt{-d}}\sinh(\sqrt{-d})\,F,\quad d<0\end{cases}$$
$$=M'=\tfrac{t'}2I+F'$$
The trace of this is
$$t'=\text{tr}(\exp(M))=\exp(\tfrac t2)\cdot\begin{cases}2\cos(\sqrt d),\quad d>0\\2,\quad d=0\\2\cosh(\sqrt{-d}),\quad d<0\end{cases}$$
and the trace-free part is
$$F'=\text{tf}(\exp(M))=\exp(\tfrac t2)\cdot\begin{cases}\frac{1}{\sqrt d}\sin(\sqrt d)\,F,\quad d>0\\F,\quad d=0\\\frac{1}{\sqrt{-d}}\sinh(\sqrt{-d})\,F,\quad d<0\end{cases}$$
which has determinant
$$d'=\det(F')=\exp(t)\cdot\begin{cases}\sin^2(\sqrt d),\quad d>0\\0,\quad d=0\\-\sinh^2(\sqrt{-d}),\quad d<0\end{cases}$$
The important part is that if $d'<0$ (or $d'=0$ and $F'\neq0$) then $t'>0$.
We can take any matrix $M$ with $d\leq0$, and multiply the two exponentials $\exp(M)=M'$ and $\exp\left(\begin{bmatrix}0&-\pi\\\pi&0\end{bmatrix}\right)=-I$, to get another matrix $M''=-M'$. This has $d''=d'\leq0$ and $F''=-F'$ but $t''=-t'<0$, so $M''$ cannot be a single exponential.
For example:
$$M''=\exp\left(\begin{bmatrix}0&-\pi\\\pi&0\end{bmatrix}\right)\exp\left(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\right)$$
$$=\begin{bmatrix}-e&0\\0&-1/e\end{bmatrix}$$
$$t''=\text{tr}(M'')=-e-1/e<0$$
$$F''=\text{tf}(M'')=\frac{-e+1/e}{2}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$
$$d''=\det(F'')=-\left(\frac{-e+1/e}{2}\right)^2<0$$
Thus, the image of $\exp$ is not a group.
A: This is not true. For example,
$$\exp \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} =\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix} \ \text{and}\ \exp \begin{bmatrix} 0&\pi \\ -\pi&0 \end{bmatrix} =\begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix}.$$
However,
$$\exp \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} =\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix}\begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix} = \begin{bmatrix} -1&-1 \\ 0&-1 \end{bmatrix}$$
is not the exponential of any real matrix.
The point is that the BCH formula gives a formula for $\exp(X) \exp(Y)$ as the exponential of a formal sum, but that sum may or may not be convergent, and the particular matrices in my example are outside the radius of convergence.
See the Wikipedia article on the BCH formula.
