# Group theoretic lemma about the extension of homomorphisms of profinite groups

I have a question about a group-theoretic lemma proven in Galois Groups and Fundamental Groups by Tamas Szamuely. Suppose we have a profinite group $\Gamma$, a closed normal subgroup $N \subset \Gamma$ and a finite group $G$. Let $n \in N, \sigma \in \Gamma, g \in G$ . We have a left action on $Hom(N,G)$ (the set of continuous homomorphisms $N \rightarrow G$) of $G$ by conjugation, i.e. $g \cdot \phi(n) = g \phi(n) g^{-1}$ and a right action of $\Gamma$ by $(\phi \cdot \sigma)(n) = \phi(\sigma n \sigma^{-1})$. This two actions commute, $g \cdot (\phi\cdot \sigma) = (g \cdot \phi) \cdot \sigma$.

Now, in this book there's a lemma saying that if $S \subset Hom(N,G)$ is stable under the actions of $G$ and $\Gamma$ and $G$ acts freely and transitively on $S$, then every $\phi\in S$ extends to a continuous homomorphism $\widehat{\phi}:\Gamma \rightarrow G$.

The author proves this by making a homomorphism by noting that $\forall \sigma \in \Gamma$ there exists a unique $g_\sigma \in G$ such that $\phi(\sigma n \sigma^{-1}) = g_\sigma \phi(n) g_{\sigma^{-1}}$ due to the fact that $S$ is stable under $G$ and $G$ acts freely and transitively on $S$. Naturally he puts $\widehat{\phi}(\sigma) = g_\sigma$ and he proves that this is a continuous extension of $\phi$. Now, I understand that it's an extension of $\phi$, but I don't understand why it's continuous. The author claims that it's enough to prove that $\ker(\widehat{\phi})$ is open (which I understand). He then goes on saying that $\ker(\widehat{\phi})$ consists of those $\sigma \in \Gamma$ that fix $\phi$ (which I also understand), but then he says that this is open due to the continuity of $\phi$. Why is that ?

Any help would be appreciated.

Edit: I know that if I can prove that the right action of $\Gamma$ on $Hom(N,G)$ (with the discrete topology) is continuous, the continuity of $\widehat{\phi}$ follows since $ker(\widehat{\phi}) = Stab(\phi)$ and this is open iff the action of $\Gamma$ is continuous. So my question boils down to proving that this action is continuous.

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What topological properties of profinite groups you know? –  Berci Nov 10 '12 at 11:36
I know that profinite groups are compact, totally disconnected, Hausdorff and that $\Gamma/N$ is a profinite group since $N$ is closed. I know that if I can prove that the action of $\Gamma$ on $Hom(N,G)$ is continuous, it is proven, but I'm not able to prove this. –  KevinDL Nov 10 '12 at 11:39
Aha. Totally disconnected. Sounds good. The stabilizer of $\phi$ is closed: $\{\sigma\in\Gamma \mid \phi\cdot\sigma =\phi \}$, because the action $Hom(N,G)\times\Gamma \to Hom(N,G)$ is continuous. –  Berci Nov 10 '12 at 11:47
But that's the problem: I'm not able to proof that the action $Hom(N,G)\times \Gamma \rightarrow Hom(N,G)$ is continuous. –  KevinDL Nov 10 '12 at 11:49
Ah, I see. How would it not be continuous...? What is the topology on $Hom(N,G)$? –  Berci Nov 10 '12 at 11:57