Uniqueness of expression of a Lie group element Just take the SU(2) group as an example. The three generators are $J_z$, $J_+$, and $J_-$. 
For an element $ g $, sometimes we want to express it as 
$$ g = e^{i a J_+} e^{i b J_z} e^{i c J_-} . $$
The problem is, for given $g$, are the parameters $a, b,c $ uniquely determined?
How about a general Lie group? 
 A: The exponentials in your factorization are uniquely determined if they exist, but not the exponents themselves. 
For related factorizations in general Lie groups see http://en.wikipedia.org/wiki/Lie_group_decompositions .
A: Comment to the question (v1): No, such decomposition is in general not unique. E.g. the unit element  ${\bf 1}_{2\times 2}\in SU(2)$ can be written with parameters $b\in 4\pi\mathbb{Z}$ and $a=0=c$.
A: Let $\mathfrak{G}$ be a Lie group of finite dimension $N$ with Lie algebra $\mathfrak{g}=\mathrm{Lie}(\mathfrak{G})$. 
Suppose further that the vectors $\{\hat{X}_j\}_{j=1}^N\subset \mathfrak{g}$ span the Lie algebra $\mathfrak{g}$ (i.e. are a $\mathfrak{g}$-basis).
Then there is some neighborhood $\mathcal{N}\subset\mathfrak{G}$ of the identity in $\mathfrak{G}$ such that for every member $\gamma\in\mathcal{N}$ we can find (in general nonunique) $\tau_j\in\mathbb{R}$ i.e.
$$\gamma=\prod\limits_{j=1}^N e^{\tau_j\,\hat{X}_j}\tag{1}$$
This is the situation in your question, where $i\,J_z,\,i\,J_\pm$ span $\mathfrak{su}(2)$. 
Indeed there is an open neighborhood $\mathcal{U}\subset\mathbb{R}^N$ of $\mathbf{0}\in\mathbb{R}^N$ in $\mathbb{R}^N$ such that the function:
$$\sigma:\mathcal{U}\to\mathcal{N}; \sigma((\tau_j)_{j=1}^N) = \prod\limits_{j=1}^N e^{\tau_j\,\hat{X}_j}\tag{2}$$
is bijective. The $\tau_j$ are then called canonical co-ordinates of the second kind for the neighborhood $\mathcal{N}$: they are the unique co-ordinates in the neighborhood $\mathcal{U}$.
Now this does not rule out  points $(\tau^\prime_j)_{j=1}^N$ outside the neighbourhood $\mathcal{U}$ such that:
$$\gamma=\prod\limits_{j=1}^N e^{\tau_j\,\hat{X}_j}=\prod\limits_{j=1}^N e^{\tau^\prime_j\,\hat{X}_j}\tag{3}$$
and this explains Qmechanic's example that any $b\in 4\pi\mathbb{Z}$ and $a=0=c$ will give you a product equal to the identity.
It follows from (2) that every member of the identity connected component of $\mathfrak{G}$ can be represented as a finite product of the form:
$$\prod\limits_{k=1}^M\,\left(\prod\limits_{j=1}^N e^{\tau_{j\,k}\,\hat{X}_j}\right)\tag{4}$$
i.e. an $M$-fold product of the canonical co-ordinate product. In a compact group, there is a strict upper limit, dependent on the exact $\mathfrak{g}$-basis chosen, for $M$, the number of multiplicands needed. In a noncompact group there is no such bound. Thus there are many group members (outside $\mathcal{N}$) in the identity component of $\mathfrak{G}$ which cannot be given canonical co-ordinates and indeed require a bigger product of the form (4). If the group member in question lies outside the identity connected component, then no product of the form in (4) will realize it. $SU(2)$, however, is connected (indeed simply connected) so you don't have this problem here. Moreover, in the special case of $SU(2)$ for the special case of the basis $\{J_z,\,J_+,\,J_-\}$ you have chosen, the function defined by (2), with three product members alone, is surjective onto $SU(2)$. It is not, however, injective, as we have already noted with Qmechanic's example.
One of the comments stated that the failure of the products (3) and (4) to realize group members was owing to the exponential map's lack of surjectivity. This is not right. In the general Lie group case, there are many group members (outside $\mathcal{N}$) in the identity component of $\mathfrak{G}$ for which there is no solution to (1), as we have noted. This is quite irrespective of whether the exponential is surjective. In a noncompact group, all members of the identity connected component can always be represented by a product of the form (4), its simply that there can be certain elements, say $\zeta$, in the identity component that are not the exponentials of any lone Lie algebra member, i.e. $\zeta\not\in\exp(\mathfrak{g})$ and there is no $X\in\mathfrak{g}$ such that $e^X=\zeta$.
The exponential map is surjective in $SU(2)$, indeed in $SU(N)$. This is easy to see because all unitary matrices are normal, thus $\gamma\in SU(N)$ unitarily diagonalizable as $\gamma=T\,\exp(i\,\mathrm{diag}(\phi_1,\,\phi_2,\,\cdots))\,T^\dagger$, where $\phi_j\in\mathbb{R}$ and so $X=i\,T\,\mathrm{diag}(\phi_1,\,\phi_2,\,\cdots)\,T^\dagger\in\mathfrak{su}(N)$ is such that $\gamma=e^X$.
The exponential map is surjective for all compact, connected Lie groups. As our example shows, though, it is not injective on the whole group, only on a restricted neighborhood of the identity.
The exponential map fails to be surjective on some noncompact connected groups: it is surjective for $SE(2)$, the proper 2D Euclidean group (probably for $SE(N)$ as well, but I haven't computed this one) but it fails to be surjective in $SL(2,\,\mathbb{R})$ and $SL(2,\,\mathbb{C})$. the example:
$$\left(\begin{array}{cc}-1&\alpha\\0&-1\end{array}\right)\in SL(2,\,\mathbb{R})\subset SL(2,\,\mathbb{C})\tag{5}$$
where $\alpha\in\mathbb{R}$ is the standard example of a Lie group element which is not the exponential of any element of $\mathfrak{sl}(2,\,\mathbb{R})$.
Naturally, $\exp$ is not surjective on any non-connected Lie group (with more than one connected component).
It is still an open question of Lie theory as to what the exact conditions for surjectivity of the exponential are.
A: In fact it is often the case that calculations simplify when the range of parameters is such that some elements are not uniquely represented.  A necessary condition for the parametrization to be unique is that the integration over the range of parameters produce the volume of the group (the volume is not always known accurately.)
