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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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    You should probably mention what $K$, $\Gamma$, and $\operatorname{mod}_K$ are.2011-07-20
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    @Dylan Moreland: Thanks for your remind, but, in view of that I did mention the book, I think this is not too big a problem, is it?2011-07-20
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    It's not too big a problem, if you're happy to do without the help of those of us who do not have that book handy. No one is going to run off to the library to borrow it, just to answer your question. You were lucky this time, as Dylan had the book (and saw your question, and was in a mood to answer it); next time, you may not be so lucky.2011-07-20
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    @awllower: It depends. I would advice you to make your question as self-contained as possible. Many potential answerers may know the subject, but not the specific book (even if the book is a 'classic' as in your case). Also many may have a copy, but it may be in the office, at home, buried under a pile of otherwork, borrowed by a colleague/student et cetera. You're hurting your chances of more answers (and upvotes, if that's a concern) by not doing that.2011-07-20
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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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    Sorry, but is q in $R_{+}^{*}$?2011-07-20
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    Also, as an unexperienced reader, I might need more than started, sorry about that. From the answer I obtain that, if the equality is not true, there is an open interval, within the two sides of the inequality, having the intersection with $\Gamma$ ={$\eta$}, but then I not know what to do to derive a contradiction; in any case, thank you very much!!2011-07-20
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    @awllower No need to be sorry. It might help to notice that $\eta = \gamma^{n + 1}$, which is what you want, is equivalent to $\eta/\gamma^n = \gamma$.2011-07-20
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    Ah, thank you very much; but why does this have anything to do with the discreteness? For some reason I cannot call you, sorry @Dylan.2011-07-20
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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