# 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 $\left< a \right>= \left< b \right>$ for two elements $a, b \in G$ then $I (a) = I (b)$ , where for any element $g
\in G, I (g) = |\{s \in S : g s = s \}|$
.
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 $a \in B$ and $b \in A$.
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), $N \cdot ord(A) = N \cdot ord(B) = \sum_{a \in A} I(a) = \sum_{b \in B} I(b)$.
By the definition of I(g), if $as = s$ then $a^n s = a^{n-1} s = ... = s$. Else, $as \neq s$ then $a^n s \neq a^{n-1} s \neq ... \neq s$. Therefore, we are sure that if $as = s$ then $bs = s$ since $b = a^z$ for some $z \in \mathbb Z$.
Is that correct?

2. Originally Posted by vincisonfire

Let G be a ﬁnite group acting on a ﬁnite set. Prove that if $\left< a \right>= \left< b \right>$ for two elements $a, b \in G$ then $I (a) = I (b)$ , where for any element $g
\in G, I (g) = |\{s \in S : g s = s \}|$
.
we have: $a=b^i$ for some $i$ coprime with $o(b),$ because $o(a)=o(b).$ so if $b^{ki}s=s, \ \forall k \in \mathbb{Z},$ then: $bs=s.$ thus: $as=s \Longleftrightarrow b^is=s \Longleftrightarrow \forall k \in \mathbb{Z}: \ b^{ki}s=s \Longleftrightarrow bs=s. \ \ \Box$

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