# Profinite group power map

I am working through some exercises in Neukirch's Algebraic number theory and i need some hints for exercise 1 pg 274, it goes as follows:

Let $G$ be a profinite group, show that we can extend the power map $G \times \mathbb{Z} \rightarrow G$, $(\theta,a) \mapsto \theta^{a}$, to a continuous map $G \times \hat{\mathbb{Z}} \rightarrow G$, $(\theta,a) \mapsto \theta^{a}$, and that $\theta^{ab} = (\theta^{a})^{b}$, $\theta^{a+b}=\theta^{a}\theta^{b}$ if $G$ is abelian.

Now my idea to extend the map was the following, since $G$ is profinite we have $G \cong \varprojlim G/N_{i}$ where the limit runs over the $open$ normal subgroups $N_{i}$ of $G$ , and we know $\hat{\mathbb{Z}} \cong \varprojlim \frac{\mathbb{Z}}{n\mathbb{Z}}$, so let $\theta \in G$, and define $\theta_{i} = \theta \mod N_{i}$, and for $a \in \hat{\mathbb{Z}}$ do a similar thing, then I defined $\theta^{a} = (\theta_{i}^{a_{i}})_i$, now im not sure if this is correct or not, and even if it is right i am not quite sure how to prove its continuous, this is where I need some hints.

Thank you

Correction: Actually, thinking about it a but more I think my way of extending the map might not work, beacuse i dont know that every profinite has enough normal subgroups in order to make the definition work, maybe I shoud define $\theta^{a} = \prod \theta^{a_{i}}$, but Im not sure how to prove its continuous.

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Well, we have $G\times \mathbb Z\subset G\times\hat{\mathbb Z}$ is a dense subset. So how about extending by continuity? –  M Turgeon Sep 25 '12 at 15:30
One minor point: $G=\varprojlim G/N$ where $N$ runs over the open normal subgroups of $G$. –  Keenan Kidwell Sep 25 '12 at 15:48
@M Turgeon: so by extending by continuity do you mean taking the limit? im not quite sure what you mean by exten by continuity. –  user39947 Sep 25 '12 at 16:02
@ Kidwell: you're right thank you –  user39947 Sep 25 '12 at 16:02
@user39947 Every element of $G\times\hat{\mathbb Z}$ is a limit point of $G\times\mathbb Z$, so to define your map $\tilde{f}:G\times\hat{\mathbb Z}\to G$ at a point $x$, take a sequence $\{x_n\}\subset G\times\mathbb Z$ converging to $x$, and set $\tilde{f}(x):=\lim f(x_n)$, where $f$ is your original power map. This is continuous by definition, and it makes it very easy to check the properties. –  M Turgeon Sep 25 '12 at 18:08