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for $N > 0$, I'm trying to show Fermat's little theorem, for $3$ using the orbit stabilizer theorem:

$N^3 - N$ an element of $3\mathbb{Z}\ (3 \mod \mathbb{Z})$

Pf/ we can break it down into multiple cases

case 1: diagonal $i = j = k$ $(i,i,i)$ and orbit of this is $1 \times N$

case 2: $i$ not equal to $j$, but $j = k$

case 3: $i$ not equal to $j$ or $k$ and $j$ not equal to $k$

Can someone help me organize this and explain what's going on?

Thanks

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    Besides every answer, is a tick mark visible! So, if you find the answer appealing, you can click on that. That will mark the answer as selected by you! The answerer gets +15, you get +2, the question stands solved!2012-01-24
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    In order to use the Orbit-Stabilizer Theorem, you need to have a group acting on a set. It is unclear to me what is your set and what is the group you are using to act on it. Could you perhaps say what these are explicitly?2012-01-24
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    it looks like my set is N^3 - N, and the group is 3 mod Z2012-01-24
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    @BuddyHolly: "It looks like"? Are you trying to do a proof, or are you trying to understand someone else's proof? If the former, then you should have a clear idea of what you are doing (if you don't, that's your first problem). If the latter, then please quote with full context and give references. To use Orbit-Stabilizer, you need to know *what* the set, what the group, and what the action are. They have to be clear and definite, not things that "look like" they might be something.2012-01-24
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    Sorry, I should say: N^3-N is the set and the group is 3 mod Z. I am trying to do both, for if my proof may be wrong then I would have to understand someone else's way at it.2012-01-24
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    @BuddyHolly: If $N$ is a positive integer, what do you mean when you say "the set is N^3 - N"? If you consider $N$ to be the set $\{0,1,2,\ldots,N-1\}$ (or perhaps $\{1,2,\ldots,N-1\}$), then $N^3$ is the set of ordered triples with entries in that set, and $N$ is that set, so $N^3-N = N^3$. If by $N^3$ you mean the number $N$ raised to the third power, then how do you consider $N^3-N$ to be a set? And what is the **action** of your group on whatever it is that your set is supposed to be? I'll wager that most, if not all, of your difficulties stem from not really knowing what you are doing...2012-01-24
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    The action of the group is things that fix x, but I'm not understanding the analogy of the orbit stabilizer theorem in this problem2012-01-24
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    @BuddyHolly: Please read the comment before answering. "$N^3-N$" does not even make sense. "The action of the group is things that fix x" does not even make sense. You don't know what you are saying, which is probably because you don't understand what you are doing. You need to take several steps back and start over, because *you are not making any sense whatsoever*.2012-01-24

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