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As usual, as I am not quite diligent a user of this website, I cannot make sure whether or not this has already been posted, but I have tried.
In the last winter, I for the first time found the book Basic Number Theory by André Weil, which I found an amount of doubts therein; I thought this is casual for a student with barely any acquaintance with the locally compact groups then. After some time spent on this topic, even though I can solve a lot of problems right now, there are some of them refusing to get into my mind, one of which is this.

On the page 13 of the book, line 6 states that $\Gamma$ is generated by $q$, the number of elements of the residual field; however, line 16 states that it is generated by an element <1; moreover, I do not even know why it can be generated by only one element.
Here, $K$ is a locally compact field, $\Gamma$ the image of $K^{*}$ under $\bmod_{K}$, where $\bmod_{K}$ is the module function on the field, obtained from the unicity of the Haar measure. Also, q is the number of elements of $R/P$, the residual field.

Since only some hints are required here, I tag this post as a homework; if this is somehow inappropriate, please notice me.
Best thanks and regards here.

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    I apologize here, for the disregard of the situations of others. And I will complete the question.2011-07-20

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At any rate, since I have the book: I think the confusion stems from the fact that the image of this absolute value is generated by both $q$ and $q^{-1}$ in $(\mathbf{R}_{> 0}, \times)$. Note that for any cyclic group, the inverse of a generator is also a generator. That $\Gamma$ is cyclic follows because it is discrete: use this to choose a maximal element $\gamma < 1$ of $\Gamma$ as Weil does. If $0 < \eta < 1$ is an element of $\Gamma$, then there is an $n$ such that $\gamma^{n + 1} \leqq \eta < \gamma^n$. That should get you started.

Added later: Let me promote this post by Mariano, which shows that a discrete subgroup of a Hausdorff topological group is closed. This is what allows Weil to choose $\gamma$. Here's a hint in our present notation: as $\log\colon (\mathbf{R}_{>0}, \times) \to (\mathbf{R}, +)$ is an isomorphism of topological groups, it suffices to show that a discrete subgroup $A$ of $\mathbf{R}$ is closed. Show that if $x$ is a limit point of $A$, then so too is $0$.

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    @DylanMoreland let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/875/discussion-between-awllower-and-dylan-moreland)2011-07-20