Let G be a ﬁnite group acting on a ﬁnite set. Prove that if for two elements then , where for any element .
Let A be the cyclic subgroup of G generated by a and B the cyclic subgroup of G generated by b.
Since A = B, we know that and .
Let N be the number of orbit in A. It is the same number as in B.
From the Cauchy-Frobenius formula (or Burnside it seems there is some disputes about who found it first), .
By the definition of I(g), if then . Else, then . Therefore, we are sure that if then since for some .
Is that correct?