Can $A, B$ fail to commute if $e^A=e^B=e^{A+B}=id$? Consider the real $n \times n$-matrices $A$ and $B$.  

Can $A, B$ fail to commute if $e^A=e^B=e^{A+B}=id$ ?

 A: Yes, $A$ and $B$ can fail to commute. Consider
$$
A = \left[\begin{array}{@{}rrr@{}}
     0 & 1 & 0 \\
    -1 & 0 & 0 \\
     0 & 0 & 0
\end{array}\right],\qquad
B = \left[\begin{array}{@{}ccc@{}}
    0                  & -\tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\
    \tfrac{1}{2}        & 0            & 0 \\
   -\tfrac{\sqrt{3}}{2} & 0            & 0
\end{array}\right],
$$
so that
$$
A + B = \left[\begin{array}{@{}ccc@{}}
     0                  & \tfrac{1}{2} & \tfrac{\sqrt{3}}{2} \\
    -\tfrac{1}{2}        & 0           & 0 \\
    -\tfrac{\sqrt{3}}{2} & 0           & 0
\end{array}\right].
$$
Since each of $A$, $B$, and $A + B$ has Frobenius norm equal to $\sqrt{2}$, we have
$$
\exp(2\pi A) = \exp(2\pi B) = \exp\bigl(2\pi (A + B)\bigr) = I.
$$
However, it's easy to check $AB$ is not symmetric, so
$$
BA = (-B^{T})(-A^{T}) = B^{T}A^{T} = (AB)^{T} \neq AB.
$$
It follows, of course, that $2\pi A$ and $2\pi B$ don't commute, either.
A: This is wrong. If $e^{A+B}=I$, then $A+B=2n\pi i$, for all $n\in\mathbf{Z}$, i.e., $2n\pi i-A=B$. Consider $[A,B]=AB-BA=A(2n\pi i-A)-(2n\pi i-A)A=-AA+AA=0$. Hence, if $e^{A+B}=I$, then $A$ and $B$ commute.
