Let $f : G \to T$ be a continuous homomorphism from a profinite group into $T= \mathbb{R}/\mathbb{Z}$.
According to a book I am reading, $f(G)$ is totally disconnected.
Why is this true? The book doesn't explain it at all.
Continuous Image of a profinite group into $T=\mathbb{R}/\mathbb{Z}$
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profinite-groups
1 Answers
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The image is finite here is a proof
The torus has an open neighborhood $V$ of zero, which contains no non-trivial subgroups of $S^1$. Since $G$ is a profinite group, there exists an open normal subgroup $U$ of $G$ satisfying $U\subseteq \chi^{-1}(V)$. This implies $U\subseteq Ker(\chi)$. So, the map $\chi$ factors throw $G/U$, which is finite (G is compact and U open). It follows the image is also finite.