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.)
Spaces with given covering group
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algebraic-topology
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3The answer is no, in general. $\pi_1 X$ would have to contain a group isomorphic to $G$ for this to be true. – 2011-03-16
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2Ryan'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 $F$. – 2011-03-16
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0One of the above comments should be added as an answer. – 2011-03-16
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2The 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 $G$. – 2011-03-24