0
$\begingroup$

I'm stuck with the proof of the following:

Let $G,H$ be profinite groups and $f: G\rightarrow H $ be a homomorphism. Show that $f$ is continous if and only $f^{-1}(N)$ is open in $G$ for every open normal subgroup $N$ in $H$.

If $f$ is continous then the statement holds obviously.

So let $f^{-1}(N)$ be open in $G$, for all open normal subgroups $N<H $. I want to show that $f^{-1}(U)$ is open in $G$ fo every open set $U \subset H$ . Since $G$ and $H$ are profinite, they are compact totally disconnected and so every open set in H is a union of cosets of open normal subgroups. $U=\bigcup Nh $ for some $h\in H$.

But from this point on I don't know how to continue. Any hint would be nice

$\endgroup$
0
$\begingroup$

Ok, I see that I already did everything necessary, since Ng is also open and the group homomorphism is continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.