first Chern class of complex vector bundles and first Pontrjagin class of quaternionic vector bundles Let $\xi$ be a (real) vector bundle of dimension $n$. Then the first Stiefel-Whitney class 
$$
w_1(\xi)=0
$$
if and only if $\xi$ is orientable, i.e. the structure group of $\xi$ can be reduced to $SO(n)$.
Question 1: Let $\xi^\mathbb{C}$ be a complex vector bundle of dimension $n$. Then the first Chern class 
$$
c_1(\xi^\mathbb{C})=0
$$
if and only if  the structure group of $\xi^\mathbb{C}$ can be reduced to $SU(n)$? Is it true or false? Any references?
Question 2: Let $\xi^\mathbb{H}$ be a quaternion vector bundle of dimension $n$. Then the first Pontrjagin class 
$$
p_1(\xi^\mathbb{C})=0
$$
if and only if  the structure group of $\xi^\mathbb{H}$ can be reduced to $SSp(\mathbb{H},n):=\{A\in Sp(n)\mid \text{Det}(A)=1\}$? Is it true or false? Any references?
I have a further question characteristic classes of a covering space with symmetric group action
 A: The natural question also relates to understanding holonomy in Riemannian geometry  using the idea of $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, and Cayley numbers as scalars. There are 8 discussions, two for each of the four choices of scalars, whether non orientable or orientable in each context.
Orientable : $SO(n)$, $SU(n)$, $Sp(n)$, and $G_2$ where $G_2$ is the 14-dimensional exceptional Lie group and non-orientables are group structures on twisted products of the four orientables with the unit spheres in each space of scalars. Namely $O(n)$, $U(n)$, $Sp(n)\cdot Sp(1)$, and $Spin(7)$. This is a current topic of research, discussing geometries that realize the groups in this exhaustive list of holonomy groups in Riemannian geometry which are maximal, not symmetric space holonomy groups and not products.
A: I think haohaizi was looking for the following standard facts:
$BSO(n)$ is the fiber of $Bdet : BO(n) \longrightarrow BO(1)$
and $Bdet = w_1$.  This uses $BO(1) = K(Z/2,1)$.
$BSU(n)$ is the fiber of $Bdet : BU(n) \longrightarrow BU(1)$
and $Bdet = c_1$.  This uses $BU(1) = K(Z,2)$.
There is no determinant function for quaternions;  they are not commutative.
As Sullivan suggests, $Sp(n)$ is already `special', i.e., oriented.  Further,
$Sp(1) = S^3$ is not a $K(Z,3)$, so the obstruction to orientability here is not a cohomology class.
