# Fermat's little theorem for $n=3$

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

-
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
@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. –  Arturo Magidin Jan 24 '12 at 5:30

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.