TITLE: Spin structures on 7-dimensional spherical space forms
QUESTION [15 upvotes]: 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.

REPLY [4 votes]: 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 $S^7$.