Proof of the first Sylow theorem in the Artin's textbook Let $G$ be a finite group, $|G|=p^n m$. Artin is using the following nice trick to show existence of a Sylow $p$-subgroup of $G$. Let $S$ be the set of all subsets of $G$ of cardinality $p^n$. Then, $G$ acts on $S$ and $|S|= \binom{p^n m}{p^n}$ is not divisible by $p$. Then there is an orbit of order not divisible by $p$, say orbit $GU$. So
$$
|Stab(U)| |GU|=|G|=p^n m.
$$
Then he claims $|Stab(U)|=p^n$ and so $Stab(U)$ is a Sylow $p$-subgroup.
I don't understand why we could not have $m=m_1 m_2$, $m_1 >1$ and $m_2 >1$ and $|Stab(U)|=p^n m_1$?

Firstly, $p^n \nmid |GU| \Rightarrow p^n \mid |\operatorname{Stab}(U)| \Rightarrow |\operatorname{Stab}(U)| \ge p^n$, as you have correctly pointed out. But consider that $\operatorname{Stab}(U)=\{g \in G\mid gU=U\}$, so that, given $a \in U$, $\forall g \in \operatorname{Stab}(U)$ it is $ga \in U$; thence, there's a map $\psi_a\colon \operatorname{Stab}(U)  \rightarrow U$, $g \mapsto ga$, injective by right cancellation, which yields $|\operatorname{Stab}(U)|\le |U|=p^n$. Finally, we are just left with $|\operatorname{Stab}(U)|=p^n$.
