Problem :

Let G be a ﬁnite group acting on a ﬁnite set. Prove that if $\displaystyle \left< a \right>= \left< b \right> $ for two elements $\displaystyle a, b \in G $ then $\displaystyle I (a) = I (b) $ , where for any element $\displaystyle g

\in G, I (g) = |\{s \in S : g s = s \}| $.

My attempt:

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 $\displaystyle a \in B $ and $\displaystyle b \in A $.

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), $\displaystyle N \cdot ord(A) = N \cdot ord(B) = \sum_{a \in A} I(a) = \sum_{b \in B} I(b)$.

By the definition of I(g), if $\displaystyle as = s $ then $\displaystyle a^n s = a^{n-1} s = ... = s $. Else, $\displaystyle as \neq s $ then $\displaystyle a^n s \neq a^{n-1} s \neq ... \neq s $. Therefore, we are sure that if $\displaystyle as = s $ then $\displaystyle bs = s $ since $\displaystyle b = a^z $ for some $\displaystyle z \in \mathbb Z $.

Is that correct?