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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?


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Please accept top answers, improve your accept rate! If you're not satisfied with the answers you've got, ask the authors to clarify. Make it a point to learn something from the questions you post here! –  user21436 Jan 24 '12 at 2:46
I agree, still new to how to accept answers. –  Buddy Holly Jan 24 '12 at 2:49
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! –  user21436 Jan 24 '12 at 2:51
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? –  Arturo Magidin Jan 24 '12 at 4:35
it looks like my set is N^3 - N, and the group is 3 mod Z –  Buddy Holly Jan 24 '12 at 5:11
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1 Answer

Consider all vectors $(a,b,c)$ with $a$, $b$, $c \in {\mathbb Z}/2{\mathbb Z}$ and let ${\mathbb Z}/3{\mathbb Z}$ act on this set by cyclic shifts: the residue class $1 \bmod 3$ acts via $(a,b,c) \to (b,c,a)$. Now count orbits.

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What is the orbits in this case? –  Buddy Holly Jan 24 '12 at 6:10
The orbit of (1,0,0) consists of (1,0,0), (0,1,0) and (0,0,1). The only elements with orbit of length < 3 are those of (0,0,0) and (1,1,1). Thus the 2^3 vectors consist of 2 orbits of length 1 and orbits of length 3, which implies 2^3 - 2 = 0 mod 3. –  franz lemmermeyer Jan 24 '12 at 17:28
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