**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.