$\exp(A+B)$ and Baker-Campbell-Hausdorff A few years ago, I did research in quantum mechanics, specifically dealing with generalized displacement operators. In such musings, BCH lights (or gets in, depending on your viewpoint) the way. A question that struck me today was: does $\exp(A+B) = \exp(A)\exp(B)$ hold if and only if $A$ and $B$ commute? Clearly, if they do commute this is true but I have not seen anything detailing the opposite direction. Given the complexity of BCH, I would be inclined to think that it's not simple to prove if it is true. I've thought about it on my own but haven't been able to come to any sort of conclusion one way or the other.
 A: The proof is actually not that hard to follow. Granted, there are proofs which require a great deal of Mathematics to follow the proof. However, there is an excellent proof given by Eichler and is explained in Stillwell's excellent book, Naive Lie Theory, which can be found here:
Naive Lie Theory by Stillwell
What you are looking for begins at page 152. However, I recommend the whole book as a read. He is an excellent author and the book is a nice undergraduate level approach to Lie Groups/Algebras.
A: I just read this file.
The fact that $\exp(A+B) = \exp(A)\exp(B)$ is not equivalent to $AB=BA$ is well-known.
More interesting (and difficult) is: when $n\geq 3$, what is the set $\{A,B\in M_n(\mathbb{C})|\exp(A+B) = \exp(A)\exp(B)\}$. Since nobody knows the answer, this question seems to be a good subject of research.
A: Take 
$A=\begin{pmatrix}i\pi & 0 \\ 0 & - i\pi\end{pmatrix}$
and
$B=\begin{pmatrix}i\pi & 1 \\ 0 & - i\pi\end{pmatrix}$.
Then $A$ and $B$ do not commute and $\exp(A)\,\exp(B)=\exp(A+B)$.
