Product of exponential of matrices Suppose that $e^Xe^Y=e^Ye^X$. 
It is known that, in general, this equality does not imply $XY=YX$. However, an exercise in Stillwell's Naive Lie Theory Page 154, Exercise 7.6.3 asks to show that $XY=YX$ (My understanding from section 7.6 in the book is that, this exercise requires the implicit assumption that the norms of all matrices involved are small).
I would appreciate possibly different hints/solutions for doing this problem.
Edit:
Fortunately, I could add the the link to the exercise through googlebooks. Please follow the link for the precise statement of the exercise and the context in which it appears.
 A: "Small norms" may mean that the sought equality holds only approximately.
$$
e^{X} e^{Y} = (I + X + \mbox{(negligible)}) (I + Y + \mbox{(negligible)}) = I + X + Y + XY + \mbox{(negligible)}
$$
Similarly,
$$
e^{Y} e^{X} = I + X + Y + YX + \mbox{(negligible)}
$$
Since you asked only for hints, I'll stop here.
A: In terms of the commutators, you've got
$$
0=[e^X,e^Y] \\
=[1+X+\frac{1}{2}X^2+O(X^3),1+Y+\frac{1}{2}Y^2+O(Y^3)] \\
=[X,Y]+\frac{1}{2}[X^2,Y]+\frac{1}{2}[X,Y^2]+O(X^3,Y^3),
$$
which implies that either $X$ and $Y$ commute, or else the commutator of $X$ and $Y$ is $O(X^2,Y^2)$, since the remaining terms are all at least that small and need to cancel it out.  If you want to be precise about what you're proving, I believe you can show that if $[e^{cX},e^{cY}]=0$ whenever $|c| < \varepsilon$ (for some positive $\varepsilon$), then $[X,Y]=0$.  (In other words, $e^{cX}$ and $e^{cY}$ can "accidentally" commute for some values of $c$, but if they do so for arbitrarily small values of $c$, then $X$ and $Y$ must themselves commute.)
