Wikipedia (http://en.wikipedia.org/wiki/Profinite_group, Properties and Facts) says that the factor group of a profinite group $G$ by a closed normal subgroup $N$ is another profinite group. No proof or reference given and I'm trying to work this out for myself. Compactness and Hausdorffness seem easy enough but I don't know how to show that $G/N$ is totally disconnected. Thanks for any help.
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There are some details to think about in the following, but this is a hint:
You want by definition to prove that every connected components are singletons. This follows if you show that the connected component that contains the identity is a singleton. Let $C$ be this connected component containing $1$ in $G/N$. You want to show that $C= \{1\}$.
Let $$\pi : G \to G / N $$ be the projection map.
Now then, let $x\in G/N$, $x\neq 1$. You have the Hausdorff, so there is an open neighborhoos $U$ of $1$ such that $x\notin U$. So then $\pi^{-1}(U)$ is an open neighborhood of $1$ in $G$. Now $G$ is profinite, so $G$ has a neighborhood basis of $1$ consisting of compact open subgroups. Let $V$ be a compact open subgroup of $1$ in $G$ such that $V \subseteq \pi^{-1}(U)$.
Now $\pi(V)$ is open and compact in $G/N$. And $\pi(V)$ contains the connected component $C$. And $x \notin \pi(V)$.
So we have shown that given any $x\neq 1$ in $G/N$, $x$ is not contained in the connected component $C$ of $1$. So $C = \{1\}$.
Thomas
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