$S=\{(x_1,\ldots,x_p):x_i\in A\}$ 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?
Cyclic permutations
3
$\begingroup$
combinatorics
-
0Then 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. – 2012-01-13
1 Answers
2
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 $p$?
-
0For those who happen upon this, the fact used here is the Orbit-Stabilizer Theorem (and then Lagrange's Theorem is helpful) The former doesn't appear to be named in the index of Dummit and Foote which is the source of this question - exercise 3.2.9 part e – 2016-10-23