# Elements of finite order in compact abelian Lie Group

If $G$ is a compact abelian Lie group, why does the $n$th power map from $G$ to $G$ form a finite covering? I cannot see why the kernel must be finite.

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The fact that the group is a Lie group rather than some other kind of topological group obviously must be used, since there are compact abelian (Hausdorff, countably-based) topological groups with infinitely-many elements of order $n$. For example, a projective limit of $(\mathbb Z/n)^\ell$, where the transition maps $(\mathbb Z/n)^{\ell+1}\rightarrow (\mathbb Z/n)^\ell$ are projections to the first $\ell$ factors.

The salient claim is that in an abelian Lie group the elements of order dividing $n$ are a discrete (closed) subgroup. Granting this, a discrete subgroup of a compact group is finite. (Note that this is true of discrete sub groups, not all sub sets.)

One easy way to see that there is a neighborhood of $e$ containing only $e$ among elements $x$ with $x^n=e$ is by the exponential map.

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Theorem: Suppose $M$ is compact and $\pi:M\rightarrow N$ is a covering map where points are closed in $N$. Then the covering has finitely many sheets.
Proof: Pick any $p\in N$. Pick an open set $U$ with $p\in U\subseteq N$ for which $\pi^{-1}(U) = \coprod V_\alpha$ is an even covering. Then $\pi^{-1}(p)\subseteq M$ is a discrete subset of $M$ (since the $V_\alpha$s are disjoint and each contains exactly one point of $\pi^{-1}(p)$). Further, $\pi^{-1}(p)$ is closed as it's the inverse image of a closed set under a continuous function. Hence, it is compact and discrete, hence finite.
Your last claim is false: for example, $\{1/n|n \in \Bbb{N}\}$ is an infinite discrete subset of $[0,1]$. –  Chris Eagle Dec 4 '12 at 20:38