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Let's give $ord(g)$ the name $d$. Let $d_0$ be some integer.

Lemma: Let $ord (g)$ be $d$ and let $d_0$/$d$, then $ord$ ($g^{d_0}$) is d/$d_0$.

I think this is saying once we know that $d_0$/$d$ then we know d/$d_0$ in an order.

I am a self taught person and was just reading about this. I just want to know how this proof would look like.

Can I use this:

You have to show that $(g^{d_0})^{\frac{d}{d_0}}=1$ and that $(g^{d_0})^{k}\neq1$ for all $k\in\mathbb N, \text{ with } k<\frac{d}{d_0}$.
$g^d=1$ and that $g^m\neq1$ for all $m\in\mathbb N, \text{ with } m<d$.

Where do I go from here? Can someone show me?

To make this simple, I would just like to see a proof. I was reading this lemma the other day and was wondering how it would look like because it seemed something good to know.

9959
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    your question is not clear...needs better explanation and formatting – Eleven-Eleven Apr 23 '13 at 05:14
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    @9959: It came out in the comments below that you don't know the definition of $\operatorname{ord}(g)$ in the context of your question. It is thus unsurprising that you do not understand the answers you were given. I fear you have wasted a bit of everyone's time, including your own. Why don't you start by looking up this definition in your text, trying to understand it, and asking a question about that if necessary? Also, when asking a question about a text it is much more clear and helpful to identify the text. – Pete L. Clark Apr 23 '13 at 14:16
  • I am just reading this for fun. I am self taught and was just wondering how it would look like. It seemed like a good proof to know maybe for other things. – 9959 Apr 24 '13 at 04:55
  • I like learning some theorems or lemmas just to know what they are. I will be able to understand it if someone showed me like others on here that I learned. There was no text involved just a source paper. Thanks. – 9959 Apr 24 '13 at 04:58
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    This has nothing to do with [set-theory] – Andrés E. Caicedo Apr 24 '13 at 05:01
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    What is a "source paper"? Do you mean that you were reading some article/paper of some sort where the notation was used but not defined? Again, why not tell us specifically what the source is? Also, do you see that it doesn't really work out well if you ask a question in which you don't know what all the words and symbols mean? – Pete L. Clark Apr 24 '13 at 06:01
  • I do know what some of the symbols mean. I might not be that familiar with the words but I will learn them. That is how I learn. It was some article/paper I was reading the other day. I do not remember what it was though. – 9959 Apr 24 '13 at 06:13
  • Is it really a problem to show me the lemma. I like learning other things for fun. That is how I learn. – 9959 Apr 24 '13 at 06:13
  • Is it not how most of us learn? By learning things that other people did? We are not going to know these things automatically. – 9959 Apr 24 '13 at 06:19
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    @9959: The first thing I did was answer your question. You replied "Can you explain more?" Then it came out that you don't know what $\operatorname{ord}(g)$ means. You can't seem to take the hint that if you don't understand the question you're asking then you're asking the wrong question. I asked you what your source is, and you are unable/unwilling to provide it. You say you are self-taught but you are asking other people here to teach you in a very tedious way. I give up. – Pete L. Clark Apr 24 '13 at 06:35

1 Answers1

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You have to show that $(g^{d_0})^{\frac{d}{d_0}}=1$ and that $(g^{d_0})^{k}\neq1$ for all $k\in\mathbb N, \text{ with } k<\frac{d}{d_0}$.
Keep in mind that $g^d=1$ and that $g^m\neq1$ for all $m\in\mathbb N, \text{ with } m<d$.

P..
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  • @9959: Do you know what is the definition of $\text{ord}(g)$? – P.. Apr 23 '13 at 06:28
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    @9959: No. It means the smallest natural number $n\in\mathbb N$ such that $g^n=1$. Maybe you should see the definitions again. – P.. Apr 23 '13 at 08:56
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    @9959: http://math.stackexchange.com/questions/229150/if-operatornameord-ma-10-find-operatornameord-ma6/229643#229643 – lab bhattacharjee Apr 24 '13 at 14:28