Spaces with given covering group Given a topological space $X$ and a finite group $G$, can we construct a covering space $Y$ of $X$, such that the full covering transformation group is isomorphic to $G$. (We may assume some conditions, if required, such as connectedness, path-connectedness etc.)
 A: The answer is no, in general.  Since $\pi_1(Y)\rightarrow \pi_1(X)$ is injective we can identify it with its image; then the group of deck transformations is isomorphic to $N(\pi_1(Y))/\pi_1(Y)$.  (Here, if $H\leq G$, then $N(H)=\{g\in G:gHg^{-1}=H\}$ is the normalizer of $H$ in $G$.  This is the largest subgroup of $G$ which contains $H$ as a normal subgroup.  Note that if $G$ is abelian, then for any subgroup $H\leq G$ we have $N(H)=G$.)  This is a purely algebraic condition: since subgroups of $\pi_1(X)$ correspond bijectively to (based) covering spaces, if we find a suitable subgroup we can find such a covering space.
This might be helpful for insight. Let $F$ be a finite set. Then the covering spaces of $X$ with $|F|$ sheets correspond exactly to the representations $\pi_1(X)\rightarrow Aut(F)$. Basically, when you go around a loop you reattach $F$ to itself using the chosen automorphism. Then you can translate between geometric statements and algebraic ones, e.g. the covering space is connected if and only if $\pi_1(X)$ acts transitively on $F$.
