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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.)

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The answer is no, in general. $\pi_1 X$ would have to contain a group isomorphic to $G$ for this to be true. –  Ryan Budney Mar 16 '11 at 5:29
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Ryan's comment is the correct answer, but 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 $F$. –  Aaron Mazel-Gee Mar 16 '11 at 6:11
    
One of the above comments should be added as an answer. –  Sean Tilson Mar 16 '11 at 13:10
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The first comment is not correct: group of covering transformations $G$ is not a subgroup of $\pi_1 X$, rather it is the quotient of the normalizer of $\pi_1 Y$ in $\pi_1 X$ by $\pi_1 Y$. For example $G = \mathbb{Z}_n$ is the group of covering transformations of the $n$-fold cover of the circle, but $\pi_1 (S^1)$ does not contain $G$. –  Robert Bell Mar 24 '11 at 10:50

1 Answer 1

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 $F$.

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As Robert Bell pointed out in comments above, "contain a group" should be replaced by "admit a quotient". –  Matt E Apr 23 '11 at 11:59
    
Oops! Thanks so much for catching this. I'm changing the answer now. –  Aaron Mazel-Gee Apr 25 '11 at 7:31

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