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Thread: Group acting on set

  1. #1
    Senior Member vincisonfire's Avatar
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    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?
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  2. #2
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    Quote Originally Posted by vincisonfire View Post

    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: $\displaystyle as=s \Longleftrightarrow b^is=s \Longleftrightarrow \forall k \in \mathbb{Z}: \ b^{ki}s=s \Longleftrightarrow bs=s. \ \ \Box$
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  3. #3
    Senior Member vincisonfire's Avatar
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    So ... my idea was correct? But the end needed a better formulation if I understand.
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