If $\exp{X}=\exp{Y}$ then is $\exp{X}\exp{Y}=\exp(X+Y)$ true? Let $G$ be a compact and path-connected Lie group. If it happens that $\exp{X}=\exp{Y}$ for some $X,Y\in\mathfrak{g}$ , then is the following true?
$$\exp{X}\cdot\exp{Y}=\exp(X+Y)$$
If yes, how can it be proven?
 A: In general, it's not true if $X$ and $Y$ don't commute. Here's a counterexample. Let $G=SU(2)$, and define $X,Y\in \operatorname{Lie}(G)$ by
$$
X = 
\begin{pmatrix}
0 & \pi \\ -\pi & 0
\end{pmatrix}
\qquad 
Y = 
\begin{pmatrix}
0 & i  \pi \\ i\pi & 0
\end{pmatrix}
.
$$
Then $\exp X = \exp Y  = -I$ and thus $\exp X\exp Y = I$. But (with the help of Mathematica)
$$
\exp(X+Y)=
\left(
\begin{array}{cc}
 \cos \left(\sqrt{2} \pi \right) & \displaystyle \frac{(1+i) \sin \left(\sqrt{2} \pi
   \right)}{\sqrt{2}} \\
 \displaystyle -\frac{(1-i) \sin \left(\sqrt{2} \pi \right)}{\sqrt{2}} & \cos \left(\sqrt{2}
   \pi \right) \\
\end{array}
\right) \ne I.
$$
Here's what's going on geometrically. The group $SU(2)$ is diffeomorphic to $S^3\subseteq \mathbb{C}^2$ under the correspondence 
$\left(\begin{smallmatrix}\phantom{-}w & z\\ -\overline z & \overline w\end{smallmatrix}\right)\leftrightarrow (w.z)$. The standard round metric on $S^3$ pulls back to a bi-invariant metric on $SU(2)$ under this correspondence. 
The matrices $X$ and $Y$ represent a pair of orthogonal tangent vectors at the identity. Because the metric is bi-invariant, the curves $t\mapsto \exp tX$ and $t\mapsto \exp tY$ are geodesics, which is to say great circles on the sphere. Because $X$ and $Y$ both have length $\pi$, their geodesics meet at time $1$ at the point $-I$ antipodal to the identity. 
However, $X+Y$ has length $\pi\sqrt{2}$, so when you follow the geodesic $t\mapsto \exp t(X+Y)$ for time $1$, you go past the antipodal point but not all the way around the sphere, and thus you end up at the point indicated above.
