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$S=\{(x_1,\ldots,x_p):x_i\in A\}$ for some set $A$ and $p$ prime. $\alpha,\beta\in S$ we say $\alpha$~$\beta$ if $\beta$ can be obtained by $\alpha$ by a cyclic permutation. This is an equivalence relation. Is it true that if an equivalence class has more than 1 element then it has exactly $p$ elements? And if it's true, could you give me a proof?

Do you know anything about group theory? –  Myself Jan 13 '12 at 1:30
yes I know group theory –  Alex M Jan 13 '12 at 1:31
Then consider the action of the cyclic group $C_p = \langle x\rangle$ on the set $A$ and try to determine the orbits. The generator acts obviously by shifting everything once to the right. –  Myself Jan 13 '12 at 1:31

1 Answer 1

up vote 1 down vote accepted

Yes, it's true. If a finite group $G$ acts on a set $S$, the cardinality of the orbit of a point $s \in S$ is equal to the index of the stabilizer of $s$ in $G$. What are the possible values of this if $G$ is cyclic of order $p$?


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