# Thread: Groups

1. ## Groups

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

2. Originally Posted by bobisback
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.