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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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    @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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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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    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 3.2012-01-24