Group Extensions and Line Bundles on $BG$ I am sure the answer to this question is well-known, but 
It is well known that the group cohomology $H^2(G,\mathbb Z)$ classifies group extensions $0\to \mathbb Z\to E\to G\to 1$ and that for a topological space $X$ elements of $H^2(X,\mathbb Z)$ are in natural bijection with complex line bundles on $X$.
My question is thus:
What is the direct correspondence between extensions of $G$ and line bundles on $BG$?
That is, given an explicit line bundle $L$ how does one construct an explicit group extension $E$ such that the two give the same cohomology class and vice versa?
 A: Suppose $G$ is finite. From the short exact sequence of trivial $G$-modules $$0\to\mathbb Z\to\mathbb C\to\mathbb C^\times\to0$$ where the map $\mathbb C\to\mathbb C^\times$ is the exponential map, using the fact that $H^p(G,\mathbb C)=0$ for $p>0$ because $\mathbb C$ is divisible, you get that $$H^2(G,\mathbb Z)\cong H^1(G,\mathbb C^\times)\cong\hom_{\mathrm{groups}}(G,\mathbb C^\times),$$ and this last group is precisely the set of one-dimensional complex $G$-modules. 
Therefore an extension of $G$ by $\mathbb Z$ determines a one dimensional complex $G$-module $V$. You can now get a line bundle on $BG$ from $V$ by constructing $EG\times_G V\to BG$.
This works for other groups $G$, like Lie groups by the same reasoning but with subtler justifications.
A: Your line bundle $L$ over $BG$ can be seen as a $G$-equivariant line bundle over a point. That is, up to isomorphism, just a continuous group homomorphism $f:G \to \mathbb{C}^{\times}$. Try to lift $f$ along $\exp: \mathbb{C} \to \mathbb{C}^{\times}$ on an open cover of $G$. The error is a $\mathbb{Z}$-valued Cech-1-cocycle on $X$, and $E$ is the total space of the associated $\mathbb{Z}$-bundle over $X$.
A: If $L \to BG$ is the complex line bundle, take the unit sphere bundle $S^1 \to S(L) \to BG$ and take $\pi_1$.
