Small step in proving $|X|\equiv _p|X|^G$ for a $p$-group $G$ and a finite set $X$. Let $G$ be a $p$-group acting on the finite set $X$, then $|X|\equiv _p|X|^G$ where $|X|^G$ is the set of the elements of $X$ fixed by every element of $G$.
At some point in the proof the author states "if $x_i$ is not a fixed point, then $[G:{\rm Stab}(x_i)]=p^r$ for some $r>0$." Which is not something I understand.
There does not seem to be an argument for the claim, so it seems it should be obvious. In case it is needed, the entire proof is in the fourth page of this document (Lemma $3.8$).
 A: $G$ is a p (p is prime) group, so $|G|=p^k$ for some $k \ge 1$. If $x \in X$,Stab(x) is a subgroup of $G$, and hence $[G:Stab(x)]$ must divide $|G|=p^k$. Since p is prime, the only possibility is that $[G:Stab(x)]=p^r$ for some $r \le k$.
A: There are two facts in play here. The first is Lagrange's theorem, which states that for a group G and a subgroup S, $|G|=(G:S)|S|$ The second is that the stabilizer forms a subgroup, so it and its index both divide G. $|G|=p^k$ for some $k$, so the only divisors are powers of $p$
EDIT after seeing your comment regarding an infinite $G$, note that the index of the stabilizer is equal to the orbit. Since $X$ is finite, it is clear the orbit must be finite. So each stabilizer has finite index. The intersections of the stabilizers has finite index (this is also the set of all $g$ that get mapped to the identity permutation). A subgroup of finite index contains a normal (to the whole group) subgroup of finite index. Then $G$ modulo this guy is a finite group and it can be shown that every element of the quotient has prime power order and you take it from there. (Recognize that the order of a coset needs to divide the order of all its representatives, and all its representatives are prime power order)
