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  1. #1
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    hi i have this question set me and i have no idea could someone help me please:

    Let G be a p-group for some prime p. Assume that G acts on a finite set X and let X^G={x memeber of X | g.x =g for all g memeber of G}
    show that order X^G=order X mod p

    Thanks Rich
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  2. #2
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    Quote Originally Posted by bobisback View Post
    hi i have this question set me and i have no idea could someone help me please:

    Let G be a p-group for some prime p. Assume that G acts on a finite set X and let X^G={x memeber of X | g.x = x for all g memeber of G}
    show that order X^G=order X mod p

    Thanks Rich
    i fixed the typo you made in the definition of X^G. well, the idea is exactly the same as the one that we use to prove that the center of every finite p-group is non-trivial:

    consider the partition \{Gx_1, \cdots , Gx_m \} for X. now x_j \in X^G if and only if |Gx_j|=1. thus |X|=\sum_{j=1}^m |Gx_j|=|X^G| + \sum_{x_j \notin X^G} |Gx_j|. but by the orbit-stabilizer theorem we

    have |Gx_j|=[G:G_{x_j}] and since x_j \notin X^G implies that |Gx_j| > 1 and so [G:G_{x_j}] > 1, we'll get p \mid [G:G_{x_j}] for all x_j \notin X^G and the result follows.
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