Let $G$ denote the $Spin(n)$ group with $n>4$ and let $\Gamma$ be a cyclic subgroup $G$ of a prime order $p >2$. When does the projection $G \to G/\Gamma$ induce a surjection between cohomology groups $H^3$ with integral coefficients?

Here is a solution for $G$ a compact, simple, connected, simplyconnected Lie group and $\Gamma$ a subgroup of the center of $G$. The group $H^3(G,Z)$ classifies $U(1)$gerbes over $G$. A gerbe $\mathcal{G}$ is in the image of the pullback map $$H^3(G/\Gamma,Z) \to H^3(G,Z)$$ if and only if it admits a $\Gamma$equivariant structure. Indeed, in this case one can form the quotient gerbe $\mathcal{G}'$ over $G/\Gamma$, and the pullback of $\mathcal{G}'$ is isomorphic to $\mathcal{G}$. The crucial point is that the basic gerbe $\mathcal{G}^1$, i.e. the one that represents a generator of $H^3(G,Z)=Z$, enjoys an explicit, Lietheoretical construction in the framework of bundle gerbes. The existence of $\Gamma$equivariant structures can then be checked by inspection of a certain obstruction class. This obstruction class has been computed explicitly for all possible central subgroups. For example, the obstruction for the gerbe $\mathcal{G}^k$ over $SU(n)$ (which represents $k \in H^3(SU(n),Z)$) vanishes if either $k$ is even, or $\Gamma$ is odd, or $\frac{n}{\Gamma}$ is even. That's how I came up with the comment to the question. All constructions and calculations are in: K. Gawedzki and N. Reis "Basic gerbe over non simply connected compact groups" J. Geom. Phys., 2003, 50, 2855. An overview is in Table 5.1 on page 143 of my phd thesis. EDIT: For noncentral groups one can still use the basic gerbe $\mathcal{G^1}$, but I don't know if the obstruction classes are then still accessible for calculations. 


Here's another approach using the LeraySerre spectral sequence. Using the fact that $\Gamma$ acts freely on $G$, the map $G \to G/\Gamma$ is a covering space, and so the cohomology of $G/\Gamma$ may be computed by the spectral sequence $$E_2^{s, t} = H^s(\Gamma, H^t(G)) \implies H^{s+t}(G/\Gamma).$$ Here $H^s(\Gamma, M)$ is the s'th group cohomology of $\Gamma$ with coefficients in the $\Gamma$representation $M$. We record three facts to start:
Let's compute part of the $E_2$term of the spectral sequence. Fact 2 implies that the $t=1$ and $2$ rows vanish entirely. Fact 3 implies that $H^0(G) = \mathbb{Z}$ and $H^3(G) = \mathbb{Z}$ are trivial $\Gamma$modules, so the $t=0$ and $3$ rows are the group cohomology described in Fact 1. Thus the only possible term that can contribute to $H^3(G/\Gamma)$ is $E_2^{0, 3} = H^3(G) = \mathbb{Z}$. There is, however, a possiblity of a single differential in this region of the spectral sequence, namely $$d_4: E_2^{0, 3} = \mathbb{Z} \longrightarrow E_2^{4, 0} = H^4(\Gamma, \mathbb{Z}) = \mathbb{Z} / p.$$ Therefore, $H^3(G/\Gamma)$ surjects onto $H^3(G)$ precisely when this differential $d_4= 0$. We note that if it's not 0, it is surjective; thus at worst $H^3(G/\Gamma)$ may be identified with an index $p$ subgroup of $H^3(G)$. So, how do we compute $d_4$? I claim that it's given as: $$H^3(G) \cong H^4(BG) \to H^4(B\Gamma) = H^4(\Gamma)$$ where the map is the restriction in cohomology, induced by the (inclusion) homomorphism $\Gamma \subseteq G$. This can be seen, for instance, by comparing this with the spectral sequence for the (rather dumb) fibration $G \to G/G=pt$. So a long winded answer to your question is: The map is surjective if and only if none of the $H^4(BG) = \mathbb{Z}$ is supported on $H^4(\Gamma) = \mathbb{Z} /p$. I would imagine that determining when that is the case is highly dependent upon the subgroup in question. 

