Fiber bundles of $G$-spaces So if $G$ is a topological group and $H,\ J$ are  closed subgroups such that $H\lhd J$, then the principal bundle $G/H\to G/J$ is trivial iff it has a global section. 

I have   questions about the general case:


If $H$ and $J$ are closed subgroups of $G$ such that $H&lt;J$ (not necessarily normal), then is the bundle $G/H\to G/J$, $\ \ gH\mapsto gJ$ trivial iff it has a global section?
If $G$ has a closed normal subgroup $N$ that acts transitively on the base $G/J$ and trivially on the fiber $J/H$. Is the bundle $G/H\to G/J$ trivial?
For 1, of course if the bundle is trivial it has a global section.  But the converse need not hold.

Consider the chain of subgroups $SO(2n)\subseteq SO(2n+1)\subseteq SO(2n+2)$.  Then the bundle in 1 is the unit tangent bundle $T^1S^{2n+1}\rightarrow S^{2n+1}$.  A section of this bundle is essentially a non-vanishing vector field on $S^{2n+1}$, so exists for all $n$.  On the other hand, Adams showed this bundle is trivial only for $n=0,1,3$.

For 2, I'm not sure what you're asking.  How does $N$ act on $J/H$?
To answer your first question, the fiber bundle being trivial is actually a more general result about principle fiber bundles.  The group structure doesn't matter since you essentially only need an arbitrary manifold as your base.  


  Suppose $(P,\pi, M, G)$ is a principle fiber bundle where $P\times
 G\to P$ is a right action and $\pi:P \to M$ is projection.  Then $P$
  is trivial if and only if there exists a smooth global section.


Proof: The first direction is trivial (if $P$ is trivial then it is a product bundle $P = M\times G$).  Conversely, suppose there is a smooth global section $\sigma:M\to P$.  Then we can define a diffeomorphism $\Phi:P \to M\times G$ as
$$
\Phi(p) \;\; =\;\; \left (\pi(p), \hat{g}\right )
$$
where $\hat{g} \in G$ is the unique group element that has $\sigma(\pi(p)) \cdot \hat{g} = p$.  Such a group element is unique since the action of $G$ is required to be free.  $\Phi$ is clearly smooth and also has a smooth inverse $\Phi^{-1}:M\times G\to P$ given by
$$
\Phi^{-1}(x,g) \;\; =\;\; \sigma(x)\cdot g.
$$
$$\tag*{$\Box$}$$