Lemma

Let $n = p^r m$, where m is an integer which is not a multiple of $p$. Then

Let S be a group of order $p^r$ (for example $\mathbf{Z}/p^r \mathbf{Z}$) and T a set with m elements. Write $X = S \times T$ and let E be the set of subsets of X with $p^r$ elements. Then $\mathrm{Card}(X) = n$, whence $\mathrm{Card}(E) = \binom{n}{p^r}$ (Set Theory, III, § 5, no. 8, Corollary 1 to Proposition 11). Let S operate on X by $s.(x, y) = (sx, y) (s, x \in S, y \in T)$ and consider the canonical extension of this operation to E. In the notation of no. 5, Proposition 11, the set $E^S$ is the set of orbits of X, that is the set of subsets $Y \subset X$ of the form $S \times \{ t \}, t \in T$, whence $\mathrm{Card}(E^S) = m$. By no. 5, Proposition 11,

which proves the lemma.

We now prove the theorem. Let G be a finite group and n its order; we write $n = p^r m$, where m is not a multiple of $p$. Let E be the set of subsets of G with $p^r$ elements. Then

whence, by virtue of the lemma, $\mathrm{Card}(E) \not\equiv 0 \pmod{p}$. Consider the extension to E of the operation of G on itself by left translation. There exists $X \in E$ whose orbit has non-zero cardinal mod. $p$. If $H_X$ denotes the stabilizer of X, then

$(G:H_x) \not\equiv 0 \pmod{p}$, which means that $p^r$ divides $\mathrm{Card}(H_x)$. But $H_x$ consists of the $s \in G$ such that $sX = X$; if $x \in X$, then $H_x \subset X.x^{-1}$, whence $\mathrm{Card}(H_x) \leqslant \mathrm{Card}(X) = p^r$. Hence $\mathrm{Card}(H_x) = p^r$.