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Posted

Question: Let G be a permutation group on the finite group omega. Suppose IGI = power of p, where p is a prime. Prove that IFix(G)I = IomegaI under mod p. (I am using IGI to denote the order of G)

 

I know this has something to do with orbits and stabilizers. I also know how to get as far as the all Fix(G) are inside the center. But that is where I can't figure out how to go any further.

Posted

All orbits are a power of p, agreed? do you know what the class equation is?

 

This is true even if omega were just a set, and it isn't necessary, though it may be true that the fixed points are in the centre of omega (I don't see it myself, in fact I'd go so far as to say it was trivially false: let G be the trivial group acting on any non-abelian group, then all points are fixed).

Posted

Yes, I realize the orbits are of power P, and I know that if the orbit it 1, it is a fixed point. HOwever, I really don't know how to go about proving it any further than that?

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