# Thread: Group acting on set

1. ## Group acting on set

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?

2. Originally Posted by vincisonfire

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 $\displaystyle g \in G, I (g) = |\{s \in S : g s = s \}|$.
we have: $\displaystyle a=b^i$ for some $\displaystyle i$ coprime with $\displaystyle o(b),$ because $\displaystyle o(a)=o(b).$ so if $\displaystyle b^{ki}s=s, \ \forall k \in \mathbb{Z},$ then: $\displaystyle bs=s.$ thus: $\displaystyle as=s \Longleftrightarrow b^is=s \Longleftrightarrow \forall k \in \mathbb{Z}: \ b^{ki}s=s \Longleftrightarrow bs=s. \ \ \Box$

3. So ... my idea was correct? But the end needed a better formulation if I understand.