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Math Help - 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  \left< a \right>= \left< b \right> for two elements  a, b \in G then  I (a) = I (b) , where for any element  g <br />
\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?
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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  \left< a \right>= \left< b \right> for two elements  a, b \in G then  I (a) = I (b) , where for any element  g <br />
\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
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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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