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

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