Spin structures on 7-dimensional spherical space forms Background
Let $M$ be a spin manifold and let $\Gamma$ be a finite group acting freely and isometrically on $M$ in such a way that $M/\Gamma$ is a smooth riemannian manifold.  The quotient will be spin if and only if the action of $\Gamma$ on $M$ lifts to the spin bundle.
For reasons having to do with $11 = 7 + 4$, I got interested in $M=S^7$ with the round metric.  There is a unique spin structure on $S^7$ and the spin bundle is
$$\mathrm{Spin}(7) \to \mathrm{Spin}(8) \to S^7.$$
A while back, together with one of my students, we investigated which smooth quotients $S^7/\Gamma$ are spin and how many inequivalent spin structures they admit.  This boils down to determining the isomorphic lifts of $\Gamma \subset \mathrm{SO}(8)$ to $\mathrm{Spin}(8)$.
There are lots of finite subgroups $\Gamma \subset \mathrm{SO}(8)$ acting freely on $S^7$, which are listed in Wolf's Spaces of constant curvature and to our surprise (this does not happen with $S^5$, say) we found that all quotients $S^7/\Gamma$ are spin; although they do not all have the same number of spin structures.  Our results were obtained by a case-by-case analysis, but we always remained with the sneaky suspicion that there ought to be a simple topological explanation.
Question
Is there one?  Perhaps based on the parallelizability of $S^7$?
Thanks in advance.
Edit
I'm answering Chris's questions in the first comment below.
The problem is indeed the existence of a subgroup $\Gamma' \subset \mathrm{Spin}(8)$ such that obvious square commutes:
$$\Gamma' \to \Gamma \to \mathrm{SO}(8) = \Gamma' \to \mathrm{Spin}(8) \to \mathrm{SO}(8)$$
and where the first map $\Gamma' \to \Gamma$ is an isomorphism.  This is the same as lifting $\Gamma \to \mathrm{SO}(8)$ via the spin double cover.
The simplest counterexample for $S^5$ is to take any freely acting cyclic subgroup $\Gamma \subset \mathrm{SO}(6)$ of even order.
 A: Here is a partial answer. If the order of $\Gamma$ is odd, then this is a trivial application of transfer maps. You have described your manifold as a quotient $\pi:S^7 \to M = S^7/\Gamma$, and hence $S^7$ is a covering space of $M$. The transfer map is a wrong way map in cohomology:
$ \tau^* : H^* (S^7) \to H^* (M) $
which exists for cohomology in, say, $\mathbb{Z}/2$-coefficients. The composition $\tau^* \pi^*$ is multiplication by the order of $\Gamma$, which in this case is an isomorphism when the order of $\Gamma$ is odd. But since the cohomology of $S^7$ vanishes in degrees 1 and 2, this proves that these groups also vanish for $M$ and hence $M$ has a spin structure and it is unique. 
The more interesting case is when $\Gamma$ is 2-primary. For example why does $\mathbb{R}P^7$ have a spin structure? I suspect that your intuition is spot on and that it has to do the framing of $S^7$. 
A: I would suspect triality is involved. The two spinor representations of spin(8) of spin(8) have the same dimension as the fundamental vector representation of spin(8) for all other spinor groups the representations don't have the same dimension and I believe this is related to triality. Here is an article on triality:
http://en.wikipedia.org/wiki/Triality
It has references to more material. Also see this article on SO(8):
http://en.wikipedia.org/wiki/SO(8)
