Let $\pi$ be a group and $K(\pi,1)$ the Eilenberg-MacLane space. Let $G$ be a finite group acting on $K(\pi,1)$ such that the following is a covering map
$$
K(\pi,1)\longrightarrow K(\pi,1)/G.
$$
**Question.** Is $K(\pi,1)/G$ an Eilenberg-MacLane space? Does
$$
K(\pi,1)/G=K(\pi \times G,1)?
$$



quotient space of Eilenberg-MacLane space
Suppose a group $H$, not necessarily finite, acts on an Eilenberg-MacLane space $BN$. The homotopy quotient $BN/H$ (which agrees with the ordinary quotient if the action of $H$ is free) fits into a fiber sequence


$$BN \to BN/H \to BH$$


and the long exact sequence in homotopy shows that $BN/H$ has vanishing higher homotopy. Hence it is an Eilenberg-MacLane space $BG$ for a group $G$ fitting into a short exact sequence


$$1 \to N \to G \to H \to 1$$


(determined by the action). In other words, it's an extension of $H$ by $N$. 


Every such extension arises in this way. Among them, semidirect products (extensions for which the above sequence splits) correspond to pointed actions of $H$ on $BN$, or equivalently actions of $H$ on $BN$ admitting a (homotopy) fixed point. 


For example, take $BN$ to be the configuration space of $n$ ordered points in $\mathbb{R}^2$, so that $N = P\_n$ is the pure braid group, and $H = S\_n$ acts by permutations of the points. Then $BN/H = BG$ is the configuration space of $n$ unordered points in $\mathbb{R}^2$, so that $G = B\_n$ is the usual braid group. The corresponding short exact sequence


$$1 \to P\_n \to B\_n \to S\_n \to 1$$


does not split.


